Merge branch 'main' into main

.github/workflows/restructure_files.yml

@@ -3,8 +3,6 @@ name: Restructure files
 on:
   push:
     branches: [ main ]
-  pull_request:
-    branches: [ main ]
   workflow_dispatch:
 
 jobs:

README.md

@@ -28,7 +28,7 @@
 
 所以基地里的题解都尽量用动画模拟，加深大家对题目的理解。往下看吧
 
-另外因为一个人的精力有限，所以基地目前只有 JAVA 代码，欢迎大家贡献其他版本代码。
+另外因为一个人的精力有限，所以基地目前只有 JAVA 代码，欢迎大家[贡献其他版本代码]()
 
 贡献仓库注意事项
 
@@ -43,7 +43,7 @@
 
 如果老哥觉得仓库很用心的话，麻烦大佬帮忙点个 star ，这也是我们一直更新下去的动力。
 
-感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助，
+感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助.
 
 > 另外我和几位老哥，给刚开始刷题，但是不知道从哪里开始刷的同学，整理了一份 【刷题大纲 】可以先按这个顺序刷，刷完之后应该就能入门，当然该仓库的大部分题解也是来自那个大纲。
 >
@@ -153,7 +153,7 @@
 - [【动画模拟】leetcode 27 移除元素](https://github.com/chefyuan/algorithm-base/blob/main/animation-simulation/%E6%95%B0%E7%BB%84%E7%AF%87/leetcode27%E7%A7%BB%E9%99%A4%E5%85%83%E7%B4%A0.md)
 - [【动画模拟】leetcode 209 长度最小的子数组](https://github.com/chefyuan/algorithm-base/blob/main/animation-simulation/%E6%95%B0%E7%BB%84%E7%AF%87/%E9%95%BF%E5%BA%A6%E6%9C%80%E5%B0%8F%E7%9A%84%E5%AD%90%E6%95%B0%E7%BB%84.md)
 - [【动画模拟】leetcode 141 环形链表](https://github.com/chefyuan/algorithm-base/blob/main/animation-simulation/%E9%93%BE%E8%A1%A8%E7%AF%87/leetcode141%E7%8E%AF%E5%BD%A2%E9%93%BE%E8%A1%A8.md)
-- [【动画模拟】leetcode 160 相交链表](https://github.com/chefyuan/algorithm-base/blob/main/animation-simulation/%E9%93%BE%E8%A1%A8%E7%AF%87/leetcode%E7%9B%B8%E4%BA%A4%E9%93%BE%E8%A1%A8.md)
+- [【动画模拟】剑指 offer 52 两个链表的第一个公共节点 & leetcode 160 相交链表](https://github.com/chefyuan/algorithm-base/blob/main/animation-simulation/%E9%93%BE%E8%A1%A8%E7%AF%87/%E5%89%91%E6%8C%87Offer52%E4%B8%A4%E4%B8%AA%E9%93%BE%E8%A1%A8%E7%9A%84%E7%AC%AC%E4%B8%80%E4%B8%AA%E5%85%AC%E5%85%B1%E8%8A%82%E7%82%B9.md)
 - [【动画模拟】leetcode 328 奇偶链表](https://github.com/chefyuan/algorithm-base/blob/main/animation-simulation/%E9%93%BE%E8%A1%A8%E7%AF%87/leetcode328%E5%A5%87%E5%81%B6%E9%93%BE%E8%A1%A8.md)
 
 ### 🏳‍🌈 栈和队列

animation-simulation/二分查找及其变种/leetcode 81不完全有序查找目标元素(包含重复值) .md

@@ -76,36 +76,4 @@ class Solution {
 }
 ```
 
-Go Code:
 
-```go
-func search(nums []int, target int) bool {
-    left := 0
-    right := len(nums) - 1
-    for (left <= right) {
-        mid := left+((right-left)>>1)
-        if (nums[mid] == target) {
-            return true
-        }
-        if (nums[mid] == nums[left]) {
-            left++
-            continue
-        }
-        if (nums[mid] > nums[left]) {
-            if (nums[mid] > target && target >= nums[left]) {
-                    right = mid - 1
-            } else if (target > nums[mid] || target < nums[left]) {
-                    left = mid + 1
-            }
-
-        }else if (nums[mid] < nums[left]) {
-            if (nums[mid] < target && target <= nums[right]) {
-                left = mid + 1
-            } else if (target < nums[mid] || target > nums[right]) {
-                right = mid - 1
-            }
-        }
-    }
-    return false
-}
-```

animation-simulation/二分查找及其变种/leetcode153搜索旋转数组的最小值.md

@@ -105,25 +105,3 @@ public:
 };
 ```
 
-Go Code:
-
-```go
-func findMin(nums []int) int {
-    left := 0
-    right := len(nums) - 1
-
-    for (left < right) {
-
-        if (nums[left] < nums[right]) {
-            return nums[left]
-        }
-        mid := left + ((right - left) >> 1)
-        if (nums[left] > nums[mid]) {
-            right = mid
-        } else {
-            left = mid + 1
-        }
-    }
-    return nums[left]
-}
-```

animation-simulation/二分查找及其变种/leetcode33不完全有序查找目标元素(不包含重复值).md

@@ -129,39 +129,3 @@ class Solution {
 }
 ```
 
-Go Code:
-
-```go
-func search(nums []int, target int) int {
-    // 左右指针
-    left := 0
-    right := len(nums) - 1
-    for (left <= right) {
-        mid := left + ((right - left) >> 1)
-        rightNum, leftNum, midNum := nums[right], nums[left], nums[mid]
-        if midNum == target {
-            return mid
-        }
-
-        //落在同一数组的情况，同时落在数组1 或 数组2
-        if midNum >= leftNum {
-            //target 落在 left 和 mid 之间，则移动我们的right，完全有序的一个区间内查找
-            if midNum > target && target >= leftNum {
-                right = mid - 1
-            } else {
-                // target落在另一个无序区间， 但和原数组的性质相符， 所以可以继续循环
-                left = mid + 1
-            }
-        } else {
-            //target 落在 mid 和 right 之间，则移动我们的left，完全有序的一个区间内查找
-            if midNum < target && target <= rightNum  {
-                left = mid + 1
-            } else {
-                // target落在另一个无序区间， 但和原数组的性质相符， 所以可以继续循环
-                right = mid - 1
-            }
-        }
-    }
-    return -1
-}
-```

animation-simulation/二分查找及其变种/leetcode34查找第一个位置和最后一个位置.md

@@ -169,48 +169,3 @@ class Solution {
 }
 ```
 
-Go Code:
-
-```go
-func searchRange(nums []int, target int) []int {
-    upper := upperBound(nums, target)
-    low := lowerBound(nums, target)
-    // 不存在情况
-    if (upper < low) {
-        return []int{-1, -1}
-    }
-
-    return []int{low, upper}
-}
-
-//计算下边界
-func lowerBound(nums []int, target int) int {
-    left, right := 0, len(nums) - 1
-    for (left <= right) {
-        //这里需要注意，计算mid
-        mid := left + ((right - left) >> 1)
-        if (target <= nums[mid]) {
-            //当目标值小于等于nums[mid]时，继续在左区间检索，找到第一个数
-            right = mid - 1
-
-        }else if (target > nums[mid]) {
-            //目标值大于nums[mid]时，则在右区间继续检索，找到第一个等于目标值的数
-            left = mid + 1
-        }
-    }
-    return left
-}
-//计算上边界
-func upperBound(nums []int, target int) int {
-    left, right := 0, len(nums) - 1
-    for (left <= right) {
-        mid := left + ((right - left) >> 1)
-        if (target >= nums[mid]) {
-            left = mid + 1
-        }else if (target < nums[mid]) {
-            right = mid - 1
-        }
-    }
-    return right
-}
-```

animation-simulation/二分查找及其变种/leetcode35搜索插入位置.md

@@ -1,4 +1,4 @@
-> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
+> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
 >
 > 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
 >
@@ -70,23 +70,3 @@ class Solution {
 
 Go Code:
 
-```go
-func searchInsert(nums []int, target int) int {
-    left, right := 0, len(nums) - 1
-    for (left <= right) {
-        mid := left + ((right - left) >> 1)
-        // 查询成功
-        if nums[mid] == target {
-            return mid
-        // 右区间
-        } else if nums[mid] < target {
-            left = mid + 1
-        // 左区间
-        } else {
-            right = mid - 1
-        }
-    }
-    return left
-
-}
-```

animation-simulation/二分查找及其变种/二维数组的二分查找.md

@@ -86,32 +86,3 @@ class Solution {
 }
 ```
 
-Go Code:
-
-```go
-func searchMatrix(matrix [][]int, target int) bool {
-    if len(matrix) == 0 {
-        return false
-    }
-    // 行数
-    row := len(matrix)
-    //列数
-    col := len(matrix[0])
-    // 行数 * 列数 - 1 为右指针
-    left, right := 0, row * col - 1
-    for (left <= right) {
-        mid := left + ((right - left) >> 1)
-        //将一维坐标变为二维坐标
-        rownum := mid / col
-        colnum := mid % col
-        if matrix[rownum][colnum] == target {
-            return true
-        } else if matrix[rownum][colnum] > target {
-            right = mid - 1
-        } else if matrix[rownum][colnum] < target {
-            left = mid + 1
-        }
-    }
-    return false
-}
-```

animation-simulation/二叉树/二叉树中序遍历（迭代）.md

@@ -78,4 +78,30 @@ class Solution {
 }
 ```
 
+Go Code:
+
+```go
+func inorderTraversal(root *TreeNode) []int {
+    res := []int{}
+    if root == nil {
+        return res
+    }
+    stk := []*TreeNode{}
+    cur := root
+    for len(stk) != 0 || cur != nil {
+        // 找到当前应该遍历的那个节点，并且把左子节点都入栈
+        for cur != nil {
+            stk = append(stk, cur)
+            cur = cur.Left
+        }
+        // 没有左子节点，则开始执行出栈操作
+        temp := stk[len(stk) - 1]
+        stk = stk[: len(stk) - 1]
+        res = append(res, temp.Val)
+        cur = temp.Right
+    }
+    return res
+}
+```
+
 ###

animation-simulation/二叉树/二叉树基础.md

@@ -426,6 +426,34 @@ class Solution {
 }
 ```
 
+Go Code:
+
+```go
+func levelOrder(root *TreeNode) [][]int {
+    res := [][]int{}
+    if root == nil {
+        return res
+    }
+    // 初始化队列时，记得把root节点加进去。
+    que := []*TreeNode{root}
+    for len(que) != 0 {
+        t := []int{}
+        // 这里一定要先求出来que的长度，因为在迭代的过程中，que的长度是变化的。
+        l := len(que)
+        for i := 0; i < l; i++ {
+            temp := que[0]
+            que = que[1:]
+            // 子节点不为空，就入队
+            if temp.Left != nil { que = append(que, temp.Left) }
+            if temp.Right != nil { que = append(que, temp.Right) }
+            t = append(t, temp.Val)
+        }
+        res = append(res, t)
+    }
+    return res
+}
+```
+
 时间复杂度：O（n） 空间复杂度：O（n）
 
 大家如果吃透了二叉树的层序遍历的话，可以顺手把下面几道题目解决掉，思路一致，甚至都不用拐弯

animation-simulation/二叉树/二叉树的前序遍历(栈).md

@@ -95,3 +95,27 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func preorderTraversal(root *TreeNode) []int {
+    res := []int{}
+    if root == nil {
+        return res
+    }
+    stk := []*TreeNode{root}
+    for len(stk) != 0 {
+        temp := stk[len(stk) - 1]
+        stk = stk[: len(stk) - 1]
+        if temp.Right != nil {
+            stk = append(stk, temp.Right)
+        }
+        if temp.Left != nil {
+            stk = append(stk, temp.Left)
+        }
+        res = append(res, temp.Val)
+    }
+    return res
+}
+```

animation-simulation/二叉树/二叉树的后续遍历 (迭代).md

@@ -104,6 +104,36 @@ class Solution {
 }
 ```
 
+Go Code:
+
+```go
+func postorderTraversal(root *TreeNode) []int {
+    res := []int{}
+    if root == nil {
+        return res
+    }
+    stk := []*TreeNode{}
+    cur := root
+    var pre *TreeNode
+    for len(stk) != 0 || cur != nil {
+        for cur != nil {
+            stk = append(stk, cur)
+            cur = cur.Left
+        }
+        // 这里符合本文最后的说法，使用先获取栈顶元素但是不弹出，根据栈顶元素的情况进行响应的处理。
+        temp := stk[len(stk) - 1]
+        if temp.Right == nil || temp.Right == pre {
+            stk = stk[: len(stk) - 1]
+            res = append(res, temp.Val)
+            pre = temp
+        } else {
+            cur = temp.Right
+        }
+    }
+    return res
+}
+```
+
 当然也可以修改下代码逻辑将 `cur = stack.pop()` 改成 `cur = stack.peek()`，下面再修改一两行代码也可以实现，这里这样写是方便动画模拟，大家可以随意发挥。
 
 时间复杂度 O（n）, 空间复杂度 O（n）

animation-simulation/前缀和/leetcode523连续的子数组和.md

@@ -100,3 +100,35 @@ public:
     }
 };
 ```
+
+Go Code:
+
+```go
+func checkSubarraySum(nums []int, k int) bool {
+    m := map[int]int{}
+    // 由于前缀和%k可能为0，所以需要给出没有元素的时候，索引位置，即-1
+    m[0] = -1
+    sum := 0
+    for i, num := range nums {
+        sum += num
+        key := sum % k
+        /*
+        // 题目中告诉k >= 1
+        key := sum
+        if k != 0 {
+            key = sum % k
+        }
+        */
+        if v, ok := m[key]; ok {
+            if i - v >= 2 {
+                return true
+            }
+            // 避免更新最小索引
+            continue
+        }
+        // 保存的是最小的索引
+        m[key] = i
+    }
+    return false
+}
+```

animation-simulation/前缀和/leetcode560和为K的子数组.md

@@ -182,3 +182,24 @@ public:
     }
 };
 ```
+
+Go Code:
+
+```GO
+func subarraySum(nums []int, k int) int {
+    m := map[int]int{}
+    // m存的是前缀和，没有元素的时候，和为0，且有1个子数组(空数组)满足条件，即m[0] = 1
+    m[0] = 1
+    sum := 0
+    cnt := 0
+    for _, num := range nums {
+        sum += num
+        if v, ok := m[sum - k]; ok {
+            cnt += v
+        }
+        // 更新满足前缀和的子数组数量
+        m[sum]++
+    }
+    return cnt
+}
+```

animation-simulation/前缀和/leetcode724寻找数组的中心索引.md

@@ -107,3 +107,23 @@ public:
     }
 };
 ```
+
+Go Code:
+
+```go
+func pivotIndex(nums []int) int {
+    presum := 0
+    for _, num := range nums {
+        presum += num
+    }
+    var leftsum int
+    for i, num := range nums {
+        // 比较左半和右半是否相同
+        if presum - leftsum - num == leftsum {
+            return i
+        }
+        leftsum += num
+    }
+    return -1
+}
+```

animation-simulation/前缀和/leetcode974和可被K整除的子数组.md

@@ -129,3 +129,21 @@ public:
     }
 };
 ```
+
+Go Code:
+
+```go
+func subarraysDivByK(nums []int, k int) int {
+    m := make(map[int]int)
+    cnt := 0
+    sum := 0
+    m[0] = 1
+    for _, num := range nums {
+        sum += num
+        key := (sum % k + k) % k
+        cnt += m[key]
+        m[key]++
+    }
+    return cnt
+}
+```

animation-simulation/单调队列单调栈/leetcode739每日温度.md

@@ -58,3 +58,26 @@ class Solution {
     }
 }
 ```
+
+GO Code:
+
+```go
+func dailyTemperatures(temperatures []int) []int {
+    l := len(temperatures)
+    if l == 0 {
+        return temperatures
+    }
+    stack := []int{}
+    arr   := make([]int, l)
+    for i := 0; i < l; i++ {
+        for len(stack) != 0 && temperatures[i] > temperatures[stack[len(stack) - 1]] {
+            idx := stack[len(stack) - 1]
+            arr[idx] = i - idx
+            stack = stack[: len(stack) - 1]
+        }
+        // 栈保存的是索引
+        stack = append(stack, i)
+    }
+    return arr
+}
+```

animation-simulation/单调队列单调栈/剑指offer59队列的最大值.md

@@ -89,4 +89,56 @@ class MaxQueue {
 }
 ```
 
+GO Code:
+
+```go
+type MaxQueue struct {
+    que []int	// 普通队列
+    deq []int	// 双端队列
+    size int	// que的队列长度
+}
+
+
+func Constructor() MaxQueue {
+    return MaxQueue{
+        que: []int{},
+        deq: []int{},
+    }
+}
+
+// Is_empty 表示队列是否为空
+func (mq *MaxQueue) Is_empty() bool {
+    return mq.size == 0
+}
+
+// Max_value 取最大值值，返回我们双端队列的对头即可，因为我们双端队列是单调递减的嘛
+func (mq *MaxQueue) Max_value() int {
+    if mq.Is_empty() { return -1 }
+    return mq.deq[0]
+}
+
+// Push_back 入队
+func (mq *MaxQueue) Push_back(value int)  {
+    mq.que = append(mq.que, value)
+    // 维护单调递减队列
+    for len(mq.deq) != 0 && mq.deq[len(mq.deq) - 1] < value {
+        mq.deq = mq.deq[:len(mq.deq) - 1]
+    }
+    mq.deq = append(mq.deq, value)
+    mq.size++
+}
+
+// Pop_front 弹出队列头元素，并且返回其值。
+func (mq *MaxQueue) Pop_front() int {
+    if mq.Is_empty() { return -1 }
+    ans := mq.que[0]
+    mq.que = mq.que[1:]
+    if mq.deq[0] == ans {
+        mq.deq = mq.deq[1:]
+    }
+    mq.size--
+    return ans
+}
+```
+
 ###

animation-simulation/单调队列单调栈/接雨水.md

@@ -117,4 +117,35 @@ class Solution {
 }
 ```
 
+GO Code:
+
+```go
+func trap(height []int) int {
+    stack := []int{}
+    water := 0
+    // 最左边部分不会接雨水，左边持续升高时，stack都会弹出所有元素。
+    for i := 0; i< len(height); i++ {
+        for len(stack) != 0 && height[i] > height[stack[len(stack) - 1]] {
+            popnum := stack[len(stack) - 1]
+            // 出现相同高度的情况（其实也可以不用处理，如果不处理，相同高度时后面的hig为0，会产生很多无效的计算）
+            for len(stack) != 0 && height[popnum] == height[stack[len(stack) - 1]] {
+                stack = stack[:len(stack) - 1]
+            }
+            if len(stack) == 0 { break }
+            le, ri := stack[len(stack) - 1], i
+            hig := min(height[ri], height[le]) - height[popnum]
+            wid := ri - le - 1
+            water += wid * hig
+        }
+        stack = append(stack, i)
+    }
+    return water
+}
+
+func min(a, b int) int {
+    if a < b { return a }
+    return b
+}
+```
+
 ###

animation-simulation/单调队列单调栈/最小栈.md

@@ -73,4 +73,48 @@ class MinStack {
 }
 ```
 
+GO Code:
+
+```go
+type MinStack struct {
+    stack  []int
+    minStk []int
+}
+
+/** initialize your data structure here. */
+func Constructor() MinStack {
+    return MinStack{
+        stack:  []int{},
+        minStk: []int{},
+    }
+}
+
+// Push 入栈，如果插入值，当前插入值小于栈顶元素，则入栈，栈顶元素保存的则为当前栈的最小元素
+func (m *MinStack) Push(x int)  {
+    m.stack = append(m.stack, x)
+    if len(m.minStk) == 0 || m.minStk[len(m.minStk) - 1] >= x {
+        m.minStk = append(m.minStk, x)
+    }
+}
+
+// Pop 出栈，如果stack出栈等于minStk栈顶元素，则说明此时栈内的最小元素改变了。
+func (m *MinStack) Pop()  {
+    temp := m.stack[len(m.stack) - 1]
+    m.stack = m.stack[: len(m.stack) - 1]
+    if temp == m.minStk[len(m.minStk) - 1] {
+        m.minStk = m.minStk[: len(m.minStk) - 1]
+    }
+}
+
+// Top stack的栈顶元素
+func (m *MinStack) Top() int {
+    return  m.stack[len(m.stack) - 1]
+}
+
+// GetMin minStk的栈顶元素
+func (m *MinStack) GetMin() int {
+    return  m.minStk[len(m.minStk) - 1]
+}
+```
+
 ###

animation-simulation/单调队列单调栈/滑动窗口的最大值.md

@@ -72,3 +72,37 @@ class Solution {
     }
 }
 ```
+
+GO Code:
+
+```go
+func maxSlidingWindow(nums []int, k int) []int {
+    l := len(nums)
+    if l == 0 {
+        return nums
+    }
+
+    arr   := []int{}
+    // 维护一个单调递减的双向队列
+    deque := []int{}
+    for i := 0; i < k; i++ {
+        for len(deque) != 0 && deque[len(deque) - 1] < nums[i] {
+            deque = deque[:len(deque) - 1]
+        }
+        deque = append(deque, nums[i])
+    }
+
+    arr = append(arr, deque[0])
+    for i := k; i < l; i++ {
+        if nums[i - k] == deque[0] {
+            deque = deque[1:]
+        }
+        for len(deque) != 0 && deque[len(deque) - 1] < nums[i] {
+            deque = deque[:len(deque) - 1]
+        }
+        deque = append(deque, nums[i])
+        arr = append(arr, deque[0])
+    }
+    return arr
+}
+```

animation-simulation/数组篇/leetcode1438绝对值不超过限制的最长子数组.md

@@ -260,3 +260,46 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func longestSubarray(nums []int, limit int) int {
+    maxdeq := []int{} // 递减队列
+    mindeq := []int{} // 递增队列
+
+    length := len(nums)
+    left, right, maxwin := 0, 0, 0
+    for right < length {
+        for len(maxdeq) != 0 && maxdeq[len(maxdeq) - 1] < nums[right] {
+            maxdeq = maxdeq[: len(maxdeq) - 1]
+        }
+        maxdeq = append(maxdeq, nums[right])
+
+        for len(mindeq) != 0 && mindeq[len(mindeq) - 1] > nums[right] {
+            mindeq = mindeq[: len(mindeq) - 1]
+        }
+        mindeq = append(mindeq, nums[right])
+
+        for maxdeq[0] - mindeq[0] > limit {
+            if maxdeq[0] == nums[left] {
+                maxdeq = maxdeq[1:]
+            }
+            if mindeq[0] == nums[left] {
+                mindeq = mindeq[1:]
+            }
+            left++
+        }
+        maxwin = max(maxwin, right - left + 1)
+        right++
+    }
+    return maxwin
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```

animation-simulation/数组篇/leetcode1两数之和.md

@@ -202,3 +202,18 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func twoSum(nums []int, target int) []int {
+    m := make(map[int]int)
+    for i, num := range nums {
+        if v, ok := m[target - num]; ok {
+            return []int{v, i}
+        }
+        m[num] = i
+    }
+    return []int{}
+}
+```

animation-simulation/数组篇/leetcode219数组中重复元素2.md

@@ -222,3 +222,24 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func containsNearbyDuplicate(nums []int, k int) bool {
+    length := len(nums)
+    if length == 0 {
+        return false
+    }
+    m := map[int]int{}
+    for i := 0; i < length; i++ {
+        if v, ok := m[nums[i]]; ok {
+            if i - v <= k {
+                return true
+            }
+        }
+        m[nums[i]] = i
+    }
+    return false
+}
+```

animation-simulation/数组篇/leetcode27移除元素.md

@@ -182,3 +182,23 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func removeElement(nums []int, val int) int {
+    length := len(nums)
+    if length == 0 {
+        return 0
+    }
+    i := 0
+    for j := 0; j < length; j++ {
+        if nums[j] == val {
+            continue
+        }
+        nums[i] = nums[j]
+        i++
+    }
+    return i
+}
+```

animation-simulation/数组篇/leetcode41缺失的第一个正数.md

@@ -273,3 +273,26 @@ public:
     }
 };
 ```
+
+Go Code:
+
+```go
+func firstMissingPositive(nums []int) int {
+    length := len(nums)
+    if length == 0 { return 1 }
+    for i := 0; i < length; i++ {
+        // 将不在正确位置的元素放在正确的位置上。
+        for nums[i] > 0 && nums[i] < length + 1 && nums[i] != i + 1 && nums[i] != nums[nums[i] - 1] {
+            j := nums[i] - 1
+            nums[i], nums[j] = nums[j], nums[i]
+        }
+    }
+	// 第一个不在正确位置上的元素就是结果。
+    for i := 0; i < length; i++ {
+        if nums[i] != i + 1 {
+            return i + 1
+        }
+    }
+    return length + 1
+}
+```

animation-simulation/数组篇/leetcode485最大连续1的个数.md

@@ -207,3 +207,27 @@ public:
     }
 };
 ```
+
+Go Code:
+
+```go
+func findMaxConsecutiveOnes(nums []int) int {
+    cnt, maxCnt := 0, 0
+    for i := 0; i < len(nums); i++ {
+        if nums[i] == 1 {
+            cnt++
+        } else {
+            maxCnt = max(maxCnt, cnt)
+            cnt = 0
+        }
+    }
+    return max(maxCnt, cnt)
+}
+
+func max(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+```

animation-simulation/数组篇/leetcode54螺旋矩阵.md

@@ -180,3 +180,40 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func spiralOrder(matrix [][]int) []int {
+    res := []int{}
+    left, right := 0, len(matrix[0]) - 1
+    top, down   := 0, len(matrix) - 1
+
+    for {
+        for i := left; i <= right; i++ {
+            res = append(res, matrix[top][i])
+        }
+        top++
+        if top > down { break }
+
+        for i := top; i <= down; i++ {
+            res = append(res, matrix[i][right])
+        }
+        right--
+        if left > right { break }
+
+        for i := right; i >= left; i-- {
+            res = append(res, matrix[down][i])
+        }
+        down--
+        if top > down { break }
+
+        for i := down; i >= top; i-- {
+            res = append(res, matrix[i][left])
+        }
+        left++
+        if left > right { break }
+    }
+    return res
+}
+```

animation-simulation/数组篇/leetcode560和为K的子数组.md

@@ -217,3 +217,22 @@ public:
     }
 };
 ```
+
+Go Code:
+
+```go
+func subarraySum(nums []int, k int) int {
+    m := map[int]int{}
+    m[0] = 1
+    sum := 0
+    cnt := 0
+    for _, num := range nums {
+        sum += num
+        if v, ok := m[sum - k]; ok {
+            cnt += v
+        }
+        m[sum]++
+    }
+    return cnt
+}
+```

animation-simulation/数组篇/leetcode59螺旋矩阵2.md

@@ -360,3 +360,47 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func generateMatrix(n int) [][]int {
+    res := make([][]int, n)
+    for i := 0; i < n; i++ {
+        res[i] = make([]int, n)
+    }
+    left, right := 0, n - 1
+    top, buttom := 0, n - 1
+    size, num := n * n, 1
+    for {
+        for i := left; i <= right; i++ {
+            res[top][i] = num
+            num++
+        }
+        top++
+        if num > size { break }
+
+        for i := top; i <= buttom; i++ {
+            res[i][right] = num
+            num++
+        }
+        right--
+        if num > size { break }
+
+        for i := right; i >= left; i-- {
+            res[buttom][i] = num
+            num++
+        }
+        buttom--
+        if num > size { break }
+
+        for i := buttom; i >= top; i-- {
+            res[i][left] = num
+            num++
+        }
+        left++
+        if num > size { break }
+    }
+    return res
+}
+```

animation-simulation/数组篇/leetcode66加一.md

@@ -122,3 +122,19 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func plusOne(digits []int) []int {
+    l := len(digits)
+    for i := l - 1; i >= 0; i-- {
+        digits[i] = (digits[i] + 1) % 10
+        if digits[i] != 0 {
+            return digits
+        }
+    }
+    digits = append([]int{1}, digits...)
+    return digits
+}
+```

animation-simulation/数组篇/leetcode75颜色分类.md

@@ -149,6 +149,30 @@ class Solution {
 }
 ```
 
+Go Code:
+
+```go
+func sortColors(nums []int)  {
+    length := len(nums)
+    left, right := 0, length - 1
+    i := left
+    for i <= right {
+        if nums[i] == 2 {
+            // 把2换到最后
+            nums[i], nums[right] = nums[right], nums[i]
+            right--
+        } else if nums[i] == 0 {
+            // 把0换到最前面
+            nums[i], nums[left] = nums[left], nums[i]
+            i++
+            left++
+        } else {
+            i++
+        }
+    }
+}
+```
+
 另外我们看这段代码，有什么问题呢？那就是我们即使完全符合时，仍会交换元素，这样会大大降低我们的效率。
 
 例如：[0,0,0,1,1,1,2,2,2]
@@ -286,3 +310,27 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func sortColors(nums []int)  {
+    length := len(nums)
+    left, right := 0, length - 1
+    for i := 0; i <= right; i++ {
+        if nums[i] == 0 {
+            // 为0时，和头交换
+            nums[i], nums[left] = nums[left], nums[i]
+            left++
+        } else if nums[i] == 2 {
+            // 为2时，和尾交换
+            nums[i], nums[right] = nums[right], nums[i]
+            right--
+            // 不为1时，需要把i减回去
+            if nums[i] != 1 {
+                i--
+            }
+        }
+    }
+}
+```

animation-simulation/数组篇/剑指offer3数组中重复的数.md

@@ -174,3 +174,24 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func findRepeatNumber(nums []int) int {
+    l := len(nums)
+    if l == 0 {
+        return -1
+    }
+    for i := 0; i < l; i++ {
+        // 将nums[i]换到i的位置。
+        for nums[i] != i {
+            if nums[i] == nums[nums[i]] {
+                return nums[i]
+            }
+            nums[i], nums[nums[i]] = nums[nums[i]], nums[i]
+        }
+    }
+    return -1
+}
+```

animation-simulation/数组篇/长度最小的子数组.md

@@ -119,3 +119,33 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func minSubArrayLen(target int, nums []int) int {
+    length := len(nums)
+    winLen := length + 1
+    i := 0
+    sum := 0
+    for j := 0; j < length; j++ {
+        sum += nums[j]
+        for sum >= target {
+            winLen = min(winLen, j - i + 1)
+            sum -= nums[i]
+            i++
+        }
+    }
+    if winLen == length + 1 {
+        return 0
+    }
+    return winLen
+}
+
+func min(a, b int) int {
+    if a < b {
+        return a
+    }
+    return b
+}
+```

animation-simulation/求和问题/三数之和.md

@@ -162,3 +162,50 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func threeSum(nums []int) [][]int {
+    res := [][]int{}
+    length := len(nums)
+    if length < 3 {
+        return res
+    }
+
+    sort.Ints(nums)
+    for i := 0; i < length - 2; i++ {
+        // 去重
+        if i != 0 && nums[i] == nums[i - 1] {
+            continue
+        }
+        l, r := i + 1, length - 1
+        for l < r {
+            /*
+            // 下面两个for循环的去重也可以用下面的代码替换
+            if l != i + 1 && nums[l] == nums[l - 1] {
+                l++
+                continue
+            }
+            */
+            if nums[i] + nums[l] + nums[r] == 0 {
+                res = append(res, []int{nums[i], nums[l], nums[r]})
+                for l < r && nums[l] == nums[l + 1] {
+                    l++
+                }
+                for l < r && nums[r] == nums[r - 1] {
+                    r--
+                }
+                l++
+                r--
+            } else if nums[i] + nums[l] + nums[r] < 0 {
+                l++
+            } else {
+                r--
+            }
+
+        }
+    }
+    return res
+}
+```

animation-simulation/求和问题/两数之和.md

@@ -54,6 +54,21 @@ class Solution {
 }
 ```
 
+Go Code:
+
+```go
+func twoSum(nums []int, target int) []int {
+    m := make(map[int]int)
+    for i, num := range nums {
+        if v, ok := m[target - num]; ok {
+            return []int{v, i}
+        }
+        m[num] = i
+    }
+    return []int{}
+}
+```
+
 ### 双指针（暴力）法
 
 #### 解析

animation-simulation/求和问题/四数之和.md

@@ -95,3 +95,57 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func fourSum(nums []int, target int) [][]int {
+    res := [][]int{}
+    length := len(nums)
+    if length < 4 {
+        return res
+    }
+
+    sort.Ints(nums)
+    for i := 0; i < length - 3; i++ {
+        // 去重
+        if i != 0 && nums[i] == nums[i - 1] {
+            continue
+        }
+        for j := i + 1; j < length - 2; j++ {
+            // 去重
+            if j != i + 1 && nums[j] == nums[j - 1] {
+                continue
+            }
+            l, r := j + 1, length - 1
+            for l < r {
+                /*
+                // 下面两个for循环的去重也可以用下面的代码替换
+                if l != i + 1 && nums[l] == nums[l - 1] {
+                    l++
+                    continue
+                }
+                */
+                sum := nums[i] + nums[j] + nums[l] + nums[r]
+                if sum == target {
+                    res = append(res, []int{nums[i], nums[j], nums[l], nums[r]})
+                    for l < r && nums[l] == nums[l + 1] {
+                        l++
+                    }
+                    for l < r && nums[r] == nums[r - 1] {
+                        r--
+                    }
+                    l++
+                    r--
+                } else if sum < target {
+                    l++
+                } else {
+                    r--
+                }
+
+            }
+        }
+    }
+    return res
+}
+```

animation-simulation/求次数问题/只出现一次的数.md

@@ -131,6 +131,26 @@ class Solution:
                 return y
 ```
 
+Go Code:
+
+```go
+func singleNumber(nums []int) int {
+    if len(nums) == 1 {
+        return nums[0]
+    }
+    m := map[int]int{}
+    for _, x := range nums {
+        m[x]++
+    }
+    for k, v := range m {
+        if v == 1 {
+            return k
+        }
+    }
+    return 0
+}
+```
+
 ### 排序搜索法
 
 #### 解析
@@ -208,6 +228,25 @@ class Solution:
         return nums[len(nums) - 1]
 ```
 
+Go Code:
+
+```go
+func singleNumber(nums []int) int {
+    if len(nums) == 1 {
+        return nums[0]
+    }
+    sort.Ints(nums)
+    for i := 1; i < len(nums) - 1; i+=2 {
+        if nums[i] == nums[i - 1] {
+            continue
+        } else {
+            return nums[i - 1]
+        }
+    }
+    return nums[len(nums) - 1]
+}
+```
+
 ### HashSet
 
 #### 解析
@@ -565,4 +604,17 @@ class Solution:
         return reduce(lambda num, x: num ^ x, nums, 0)
 ```
 
+Go Code:
+
+```go
+func singleNumber(nums []int) int {
+    res := 0
+    for _, x := range nums {
+        res ^= x
+    }
+    return res
+
+}
+```
+
 本题一共介绍了 6 种解题方法，肯定还有别的方法，欢迎大家讨论。大家可以在做题的时候一题多解。这样能大大提高自己解题能力。下面我们来看一下这些方法如何应用到其他题目上。

animation-simulation/求次数问题/只出现一次的数2.md

@@ -216,6 +216,27 @@ class Solution:
         return res
 ```
 
+Go Code:
+
+```go
+func singleNumber(nums []int) int {
+    res := 0
+    // Go语言中，int占32位以上
+    for i := 0; i < 64; i++ {
+        cnt := 0
+        for j := 0; j < len(nums); j++ {
+            if (nums[j] >> i & 1) == 1 {
+                cnt++
+            }
+        }
+        if cnt % 3 != 0{
+            res = res | 1 << i
+        }
+    }
+    return res
+}
+```
+
 我们来解析一下我们的代码：
 
 > **<<** 左移运算符：运算数的各二进位全部左移若干位，由 **<<** 右边的数字指定了移动的位数，高位丢弃，低位补 0。

animation-simulation/求次数问题/只出现一次的数3.md

@@ -263,3 +263,26 @@ class Solution:
                 arr[1] ^= y
         return arr
 ```
+
+Go Code:
+
+```go
+func singleNumber(nums []int) []int {
+    temp := 0
+    for _, x := range nums {
+        temp ^= x
+    }
+    // 保留最后那个1，为了区分两个数
+    group := temp & (^temp + 1)
+
+    res := make([]int, 2)
+    for _, x := range nums {
+        if group & x == 0{
+            res[0] ^= x
+        } else {
+            res[1] ^= x
+        }
+    }
+    return res
+}
+```

animation-simulation/链表篇/234. 回文链表.md

@@ -173,6 +173,30 @@ class Solution {
 }
 ```
 
+Go Code:
+
+```go
+func isPalindrome(head *ListNode) bool {
+    // 将节点中的值按顺序放在arr中。
+    arr := []int{}
+    node := head
+    for node != nil {
+        arr = append(arr, node.Val)
+        node = node.Next
+    }
+    // 双指针判断是否为回文
+    l, r := 0, len(arr) - 1
+    for l < r {
+        if arr[l] != arr[r] {
+            return false
+        }
+        l++
+        r--
+    }
+    return true
+}
+```
+
 这个方法可以直接通过，但是这个方法需要辅助数组，那我们还有其他更好的方法吗？
 
 **双指针翻转链表法**
@@ -457,3 +481,52 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func isPalindrome(head *ListNode) bool {
+    if head == nil || head.Next == nil {
+        return true
+    }
+
+    midNode := searchMidNode(head)
+    backHalf := reverse(midNode.Next)
+
+    // 判断左右两边是否一样（回文）
+    p1, p2 := head, backHalf
+    for p2 != nil {
+        if p1.Val != p2.Val {
+            midNode.Next = reverse(backHalf)
+            return false
+        }
+        p1 = p1.Next
+        p2 = p2.Next
+    }
+    // 不破坏原来的数据
+    midNode.Next = reverse(backHalf)
+    return true
+}
+
+// searchMidNode 求中间的节点
+func searchMidNode(head *ListNode) *ListNode {
+    fast, slow := head, head
+    for fast.Next != nil && fast.Next.Next != nil {
+        fast = fast.Next.Next
+        slow = slow.Next
+    }
+    return slow
+}
+
+// reverse 反转链表
+func reverse(node *ListNode) *ListNode {
+    var pre *ListNode
+    for node != nil {
+        nxt := node.Next
+        node.Next = pre
+        pre = node
+        node = nxt
+    }
+    return pre
+}
+```

animation-simulation/链表篇/leetcode141环形链表.md

@@ -124,3 +124,20 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func hasCycle(head *ListNode) bool {
+    if head == nil { return false }
+    s, f := head, head
+    for f != nil && f.Next != nil {
+        s = s.Next
+        f = f.Next.Next
+        if s == f {
+            return true
+        }
+    }
+    return false
+}
+```

animation-simulation/链表篇/leetcode142环形链表2.md

@@ -297,3 +297,27 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func detectCycle(head *ListNode) *ListNode {
+    if head == nil { return nil }
+    s, f := head, head
+    for f != nil && f.Next != nil {
+        s = s.Next
+        f = f.Next.Next
+        // 快慢指针相遇
+        if f == s {
+            // 快指针从头开始一步一步走，也可以用一个新的指针
+            f = head
+            for f != s {
+                f = f.Next
+                s = s.Next
+            }
+            return f
+        }
+    }
+    return nil
+}
+```

animation-simulation/链表篇/leetcode147对链表进行插入排序.md

@@ -257,3 +257,34 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func insertionSortList(head *ListNode) *ListNode {
+    if head == nil || head.Next == nil { return head }
+    root := &ListNode{
+        Next: head,
+    }
+    cur, nxt := head, head.Next
+    for nxt != nil {
+        // 有序的不需要换位置
+        if cur.Val <= nxt.Val {
+            cur = cur.Next
+            nxt = nxt.Next
+            continue
+        }
+        temp := root
+        for temp.Next.Val <= nxt.Val {
+            temp = temp.Next
+        }
+        // 此时找到合适的位置
+        cur.Next = nxt.Next
+        nxt.Next = temp.Next
+        temp.Next = nxt
+        // 继续向下
+        nxt = cur.Next
+    }
+    return root.Next
+}
+```

animation-simulation/链表篇/leetcode206反转链表.md

@@ -164,6 +164,27 @@ class Solution {
 }
 ```
 
+Go Code:
+
+```go
+func reverseList(head *ListNode) *ListNode {
+    if head == nil || head.Next == nil { return head }
+    cur := head
+    var pre *ListNode
+    for cur != nil {
+        nxt := cur.Next
+        cur.Next = pre
+        pre = cur
+        cur = nxt
+        if nxt == nil {
+            return pre
+        }
+        nxt = nxt.Next
+    }
+    return pre
+}
+```
+
 上面的迭代写法是不是搞懂啦，现在还有一种递归写法，不是特别容易理解，刚开始刷题的同学，可以只看迭代解法。
 
 **题目代码**

animation-simulation/链表篇/leetcode328奇偶链表.md

@@ -155,3 +155,23 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func oddEvenList(head *ListNode) *ListNode {
+    if head == nil || head.Next == nil {
+        return head
+    }
+    odd, even := head, head.Next
+    evenHead := even
+    for odd.Next != nil && even.Next != nil {
+        odd.Next = even.Next
+        odd = odd.Next
+        even.Next = odd.Next
+        even = even.Next
+    }
+    odd.Next = evenHead
+    return head
+}
+```

animation-simulation/链表篇/leetcode82删除排序链表中的重复元素II.md

@@ -194,3 +194,30 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func deleteDuplicates(head *ListNode) *ListNode {
+	// 新建一个头结点，他的下一个节点才是开始
+    root := &ListNode{
+        Next: head,
+    }
+    pre, cur := root, head
+    for cur != nil && cur.Next != nil {
+        if cur.Val == cur.Next.Val {
+            // 相等的话，cur就一直向后移动
+            for cur != nil && cur.Next != nil && cur.Val == cur.Next.Val {
+                cur = cur.Next
+            }
+            // 循环后移动到了最后一个相同的节点。
+            cur = cur.Next
+            pre.Next = cur
+        } else {
+            cur = cur.Next
+            pre = pre.Next
+        }
+    }
+    return root.Next
+}
+```

animation-simulation/链表篇/leetcode86分隔链表.md

@@ -187,3 +187,29 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func partition(head *ListNode, x int) *ListNode {
+    big, small := &ListNode{}, &ListNode{}
+    headBig, headSmall := big, small
+    temp := head
+    for temp != nil {
+        // 分开存
+        if temp.Val < x {
+            small.Next = temp
+            small = small.Next
+        } else {
+            big.Next = temp
+            big = big.Next
+        }
+        temp = temp.Next
+    }
+    // 最后一个节点指向nil
+    big.Next = nil
+    // 存小数的链表和存大数的连起来
+    small.Next = headBig.Next
+    return headSmall.Next
+}
+```

animation-simulation/链表篇/leetcode92反转链表2.md

@@ -263,3 +263,46 @@ class Solution {
     }
 }
 ```
+
+GoCode:
+
+```go
+func reverseBetween(head *ListNode, left int, right int) *ListNode {
+    root := &ListNode{
+        Next: head,
+    }
+    temp := root
+    i := 0
+    // left的前一个节点
+    for ; i < left - 1; i++ {
+        temp = temp.Next
+    }
+    leftNode := temp
+    // right的后一个节点
+    for ; i < right; i++ {
+        temp = temp.Next
+    }
+    rightNode := temp.Next
+    // 切断链表
+    temp.Next = nil
+    newhead := leftNode.Next
+    leftNode.Next = nil
+
+    // 反转后将3段链表接上。
+    leftNode.Next = reverse(newhead)
+    newhead.Next = rightNode
+    return root.Next
+}
+
+func reverse(head *ListNode) *ListNode {
+    var pre *ListNode
+    cur := head
+    for cur != nil {
+        temp := cur
+        cur = cur.Next
+        temp.Next = pre
+        pre = temp
+    }
+    return pre
+}
+```

animation-simulation/链表篇/剑指Offer25合并两个排序的链表.md

@@ -146,3 +146,29 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
+    root := &ListNode{}
+    node := root
+    for l1 != nil && l2 != nil {
+        if l1.Val < l2.Val {
+            node.Next = l1
+            l1 = l1.Next
+        } else {
+            node.Next = l2
+            l2 = l2.Next
+        }
+        node = node.Next
+    }
+	// node接上l1或l2剩下的节点
+    if l1 != nil {
+        node.Next = l1
+    } else {
+        node.Next = l2
+    }
+    return root.Next
+}
+```

animation-simulation/链表篇/剑指Offer52两个链表的第一个公共节点.md

@@ -265,6 +265,28 @@ class Solution {
 }
 ```
 
+Go Code:
+
+```go
+func getIntersectionNode(headA, headB *ListNode) *ListNode {
+    tempA, tempB := headA, headB
+    for tempA != tempB {
+        // 如果不为空就指针下移，为空就跳到另一链表的头部
+        if tempA == nil {
+            tempA = headB
+        } else {
+            tempA = tempA.Next
+        }
+        if tempB == nil {
+            tempB = headA
+        } else {
+            tempB = tempB.Next
+        }
+    }
+    return tempA
+}
+```
+
 好啦，链表的题目就结束啦，希望大家能有所收获，下周就要更新新的题型啦，继续坚持，肯定会有收获的。
 
 <br/>

animation-simulation/链表篇/剑指offer22倒数第k个节点.md

@@ -163,3 +163,21 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func getKthFromEnd(head *ListNode, k int) *ListNode {
+    if head == nil { return head }
+    pro, after := head, head
+    //先移动绿指针到指定位置
+    for i := 0; i < k - 1; i++ {
+        pro = pro.Next
+    }
+    for pro.Next != nil {
+        pro = pro.Next
+        after = after.Next
+    }
+    return after
+}
+```

animation-simulation/链表篇/面试题 02.03. 链表中间节点.md

@@ -135,3 +135,17 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func middleNode(head *ListNode) *ListNode {
+	// 快慢指针
+    fast, slow := head, head
+    for fast != nil && fast.Next != nil {
+        fast = fast.Next.Next
+        slow = slow.Next
+    }
+    return slow
+}
+```

animation-simulation/链表篇/面试题 02.05. 链表求和.md

@@ -267,3 +267,40 @@ class Solution {
     }
 }
 ```
+
+Go Code:
+
+```go
+func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
+    root := &ListNode{}
+    temp := root
+    // 用来保存进位值，初始化为0
+    mod := 0
+    for (l1 != nil || l2 != nil) {
+        l1num := 0
+        if l1 != nil { l1num = l1.Val }
+        l2num := 0
+        if l2 != nil { l2num = l2.Val }
+        // 将链表的值和进位值相加，得到为返回链表的值
+        sum := l1num + l2num + mod
+        // 更新进位值，例18/10=1，9/10=0
+        mod = sum / 10
+        // 新节点保存的值，18%8=2，则添加2
+        sum = sum % 10
+        newNode := &ListNode{
+            Val: sum,
+        }
+        temp.Next = newNode
+        temp = temp.Next
+        if l1 != nil { l1 = l1.Next }
+        if l2 != nil { l2 = l2.Next }
+    }
+    if mod != 0 {
+        newNode := &ListNode{
+            Val: mod,
+        }
+        temp.Next = newNode
+    }
+    return root.Next
+}
+```