为数组篇 增加 Swift 实现

animation-simulation/数组篇/leetcode1052爱生气的书店老板.md

@@ -114,5 +114,45 @@ class Solution:
             ans = max(ans, t)
         return ans
 ```
- 
 
+Swift Code
+
+```swift
+class Solution {
+    func maxSatisfied(_ customers: [Int], _ grumpy: [Int], _ minutes: Int) -> Int {
+        let len = customers.count
+        var winSum = 0, rightSum = 0, leftSum = 0
+        // 右区间的值
+        for i in minutes..<len {
+            if grumpy[i] == 0 {
+                rightSum += customers[i]
+            }
+        }
+        // 窗口的值
+        for i in 0..<minutes {
+            winSum += customers[i]
+        }
+        var maxCustomer = winSum + leftSum + rightSum
+        // 窗口左边缘
+        var left = 1, right = minutes
+        while right < len {
+            // 重新计算左区间的值，也可以用 customer 值和 grumpy 值相乘获得
+            if grumpy[left - 1] == 0 {
+                leftSum += customers[left - 1]
+            }
+            // 重新计算右区间值
+            if grumpy[right] == 0 {
+                rightSum -= customers[right]
+            }
+            // 窗口值
+            winSum = winSum - customers[left - 1] + customers[right]
+            maxCustomer = max(maxCustomer, winSum + leftSum + rightSum) // 保留最大值
+            // 移动窗口
+            left += 1
+            right += 1
+        }
+        
+        return maxCustomer
+    }
+}
+```

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

@@ -111,3 +111,152 @@ class Solution:
         return maxwin
 ```
 
+Swift Code
+
+Swift：数组模拟，超时（58 / 61 个通过测试用例）
+```swift
+class Solution {
+    func longestSubarray(_ nums: [Int], _ limit: Int) -> Int {
+        var maxQueue:[Int] = []
+        var minQueue:[Int] = []
+        let len = nums.count
+        var right = 0, left = 0, maxWin = 0
+        while right < len {
+            while !maxQueue.isEmpty && (maxQueue.last! < nums[right]) {
+                maxQueue.removeLast()
+            }
+            while !minQueue.isEmpty && (minQueue.last! > nums[right]) {
+                minQueue.removeLast()
+            }
+            maxQueue.append(nums[right])
+            minQueue.append(nums[right])
+            while (maxQueue.first! - minQueue.first!) > limit {
+                if maxQueue.first! == nums[left] {
+                    maxQueue.removeFirst()
+                }
+                if minQueue.first! == nums[left] {
+                    minQueue.removeFirst()
+                }
+                left += 1
+            }
+            maxWin = max(maxWin, right - left + 1)
+            right += 1
+        }
+        return maxWin
+    }
+}
+```
+
+Swift：使用双端队列（击败了100.00%）
+
+```swift
+class Solution {
+    func longestSubarray(_ nums: [Int], _ limit: Int) -> Int {
+        var maxQueue = Deque<Int>.init()
+        var minQueue = Deque<Int>.init()
+        let len = nums.count
+        var right = 0, left = 0, maxWin = 0
+        while right < len {
+            while !maxQueue.isEmpty && (maxQueue.peekBack()! < nums[right]) {
+                maxQueue.dequeueBack()
+            }
+            while !minQueue.isEmpty && (minQueue.peekBack()! > nums[right]) {
+                minQueue.dequeueBack()
+            }
+            maxQueue.enqueue(nums[right])
+            minQueue.enqueue(nums[right])
+            while (maxQueue.peekFront()! - minQueue.peekFront()!) > limit {
+                if maxQueue.peekFront()! == nums[left] {
+                    maxQueue.dequeue()
+                }
+                if minQueue.peekFront()! == nums[left] {
+                    minQueue.dequeue()
+                }
+                left += 1
+            }
+            maxWin = max(maxWin, right - left + 1)
+            right += 1
+        }
+        return maxWin
+    }
+
+    // 双端队列数据结构
+    public struct Deque<T> {
+        private var array: [T?]
+        private var head: Int
+        private var capacity: Int
+        private let originalCapacity: Int
+
+        public init(_ capacity: Int = 10) {
+            self.capacity = max(capacity, 1)
+            originalCapacity = self.capacity
+            array = [T?](repeating: nil, count: capacity)
+            head = capacity
+        }
+
+        public var isEmpty: Bool {
+            return count == 0
+        }
+
+        public var count: Int {
+            return array.count - head
+        }
+
+        public mutating func enqueue(_ element: T) {
+            array.append(element)
+        }
+
+        public mutating func enqueueFront(_ element: T) {
+            if head == 0 {
+            capacity *= 2
+            let emptySpace = [T?](repeating: nil, count: capacity)
+            array.insert(contentsOf: emptySpace, at: 0)
+            head = capacity
+            }
+
+            head -= 1
+            array[head] = element
+        }
+
+        public mutating func dequeue() -> T? {
+            guard head < array.count, let element = array[head] else { return nil }
+
+            array[head] = nil
+            head += 1
+
+            if capacity >= originalCapacity && head >= capacity*2 {
+            let amountToRemove = capacity + capacity/2
+            array.removeFirst(amountToRemove)
+            head -= amountToRemove
+            capacity /= 2
+            }
+            return element
+        }
+
+        public mutating func dequeueBack() -> T? {
+            if isEmpty {
+            return nil
+            } else {
+            return array.removeLast()
+            }
+        }
+
+        public func peekFront() -> T? {
+            if isEmpty {
+            return nil
+            } else {
+            return array[head]
+            }
+        }
+
+        public func peekBack() -> T? {
+            if isEmpty {
+            return nil
+            } else {
+            return array.last!
+            }
+        }
+    }
+}
+```
+

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

@@ -76,6 +76,32 @@ class Solution:
         return rearr
 ```
 
+Swift Code:
+
+```swift
+class Solution {
+    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
+        let count = nums.count
+        if count < 2 {
+            return [0]
+        }
+        
+        var rearr: [Int] = []
+        // 查询元素
+        for i in 0..<count {
+            for j in i+1..<count {
+                // 发现符合条件情况
+                if nums[i] + nums[j] == target {
+                    rearr.append(i)
+                    rearr.append(j)
+                }
+            }
+        }
+        return rearr
+    }
+}
+```
+
 **哈希表**
 
 **解析**
@@ -159,3 +185,20 @@ class Solution:
         return [0]
 ```
 
+Swift Code:
+
+```swift
+class Solution {
+    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
+        var m:[Int:Int] = [:]
+        for i in 0..<nums.count {
+            let n = nums[i]
+            if let k = m[target - n] { // 如果存在则返回
+                return [k, i]
+            }
+            m[n] = i // 不存在则存入
+        }
+        return [0]
+    }
+}
+```

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

@@ -78,6 +78,32 @@ class Solution:
         return False
 ```
 
+Swift Code
+
+```swift
+class Solution {
+    func containsNearbyDuplicate(_ nums: [Int], _ k: Int) -> Bool {
+        if nums.count == 0 {
+            return false
+        }
+        var dict:[Int:Int] = [:]
+        for i in 0..<nums.count {
+            // 如果含有
+            if let v = dict[nums[i]] {
+                // 判断是否小于K，如果小于等于则直接返回
+                let abs = abs(i - v)
+                if abs <= k {
+                    return true
+                }
+            }
+            // 更新索引，此时有两种情况，不存在，或者存在时，将后出现的索引保存
+            dict[nums[i]] = i
+        }
+        return false
+    }
+}
+```
+
 **HashSet**
 
 **解析**
@@ -140,3 +166,28 @@ class Solution:
                 s.remove(nums[i - k])
         return False
 ```        
+
+Swift Code
+
+```swift
+class Solution {
+    func containsNearbyDuplicate(_ nums: [Int], _ k: Int) -> Bool {
+        if nums.count == 0 {
+            return false
+        }
+        var set:Set<Int> = []
+        for i in 0..<nums.count {
+            // 含有该元素，返回true
+            if set.contains(nums[i]) {
+                return true
+            }
+            // 添加新元素
+            set.insert(nums[i])
+            if set.count > k {
+                set.remove(nums[i - k])
+            }
+        }
+        return false
+    }
+}
+```

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

@@ -164,3 +164,23 @@ public:
 };
 ```
 
+Swift Code
+
+```swift
+class Solution {
+    func removeElement(_ nums: inout [Int], _ val: Int) -> Int {
+        if nums.count == 0 {
+            return 0
+        }
+        var i = 0
+        for j in 0..<nums.count {
+            if nums[j] != val {
+                nums[i] = nums[j]
+                i += 1
+            }
+            
+        }
+        return i
+    }
+}
+```

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

@@ -89,6 +89,33 @@ class Solution:
         return len(res)
 ```
 
+Swift Code
+
+```swift
+class Solution {
+    func firstMissingPositive(_ nums: [Int]) -> Int {
+        if nums.count == 0 {
+            return 1
+        }
+        // 因为是返回第一个正整数，不包括 0，所以需要长度加1，细节1
+        var res:[Int] = Array.init(repeating: 0, count: nums.count + 1)
+        // 将数组元素添加到辅助数组中
+        for x in nums {
+            if x > 0 && x < res.count {
+                res[x] = x
+            }
+        }
+        // 遍历查找,发现不一样时直接返回
+        for i in 1..<res.count {
+            if res[i] != i {
+                return i
+            }
+        }
+        // 缺少最后一个，例如 1，2，3此时缺少 4 ，细节2
+        return res.count
+    }
+}
+```
 
 我们通过上面的例子了解这个解题思想，我们有没有办法不使用辅助数组完成呢？我们可以使用原地置换，直接在 nums 数组内，将值换到对应的索引处，与上个方法思路一致，只不过没有使用辅助数组，理解起来也稍微难理解一些。
 
@@ -166,3 +193,42 @@ class Solution:
         return n + 1
 ```
 
+Swift Code
+
+```swift
+class Solution {
+    func firstMissingPositive(_ nums: [Int]) -> Int {
+        var nums = nums
+        let len = nums.count
+        if len == 0 {
+            return 1
+        }
+        // 遍历数组
+        for i in 0..<len {
+            // 需要考虑指针移动情况，大于0，小于len+1，不等与i+1，
+            // 两个交换的数相等时，防止死循环
+            while nums[i] > 0 
+                  && nums[i] < len + 1 
+                  && nums[i] != i + 1
+                  && nums[i] != nums[nums[i] - 1] 
+            {
+                //nums.swapAt(i, (nums[i] - 1)) // 系统方法
+                self.swap(&nums, i, (nums[i] - 1)) // 自定义方法
+            }
+        }
+        // 遍历寻找缺失的正整数
+        for i in 0..<len {
+            if nums[i] != i + 1 {
+                return i + 1
+            }
+        }
+
+        return len + 1
+    }
+    func swap(_ nums: inout [Int], _ i: Int, _ j: Int) {
+        let temp = nums[i]
+        nums[i] = nums[j]
+        nums[j] = temp
+    }
+}
+```

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

@@ -84,6 +84,32 @@ class Solution:
         return max(maxcount, right - left)
 ```
 
+Swift Code
+
+```swift
+class Solution {
+    func findMaxConsecutiveOnes(_ nums: [Int]) -> Int {
+
+        var left = 0, right = 0, res = 0
+        let len = nums.count
+        while right < len {
+            if nums[right] == 1 {
+                right += 1
+                continue
+            }
+            // 保存最大值
+            res = max(res, right - left)
+            // 跳过 0 的情况
+            while right < len && nums[right] == 0 {
+                right += 1
+            }
+            // 同一起点继续遍历
+            left = right
+        }
+        return max(res, right - left)
+    }
+}
+```
 
 刚才的效率虽然相对高一些，但是代码不够优美，欢迎各位改进，下面我们说一下另外一种情况，一个特别容易理解的方法。
 
@@ -132,3 +158,22 @@ class Solution:
         return ans
 ```
 
+Swift Code
+
+```swift
+class Solution {
+    func findMaxConsecutiveOnes(_ nums: [Int]) -> Int {
+        let len = nums.count
+        var maxCount = 0, count = 0
+        for i in 0..<len {
+            if nums[i] == 1 {
+                count += 1
+            } else { // 这里可以改成 while
+                maxCount = max(maxCount, count)
+                count = 0
+            }
+        }        
+        return max(maxCount, count)
+    }
+}
+```

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

@@ -120,5 +120,40 @@ class Solution:
         return arr
 ```
 
+Swift Code
 
+```swift
+class Solution {
+    func spiralOrder(_ matrix: [[Int]]) -> [Int] {
+        var arr:[Int] = []
+        var left = 0, right = matrix[0].count - 1
+        var top = 0, down = matrix.count - 1
+
+        while (true) {
+            for i in left...right {
+                arr.append(matrix[top][i])
+            }
+            top += 1
+            if top > down { break }
+            for i in top...down {
+                arr.append(matrix[i][right])
+            }
+            right -= 1
+            if left > right { break}
+            for i in stride(from: right, through: left, by: -1) {
+                arr.append(matrix[down][i])
+            }
+            down -= 1
+            if top > down { break}
+            for i in stride(from: down, through: top, by: -1) {
+                arr.append(matrix[i][left])
+            }
+            left += 1
+            if left > right { break}
+        }
+
+        return arr
+    }
+}
+```
 

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

@@ -45,6 +45,8 @@ class Solution {
 
 Python3版本的代码会超时
 
+Swift版本的代码会超时
+
 下面我们我们使用前缀和的方法来解决这个题目，那么我们先来了解一下前缀和是什么东西。其实这个思想我们很早就接触过了。见下图
 
 ![](https://cdn.jsdelivr.net/gh/tan45du/github.io.phonto2@master/myphoto/微信截图_20210113193831.4wk2b9zc8vm0.png)
@@ -160,3 +162,27 @@ class Solution {
 }
 ```
 
+Swift Code
+
+```swift
+class Solution {
+    func subarraySum(_ nums: [Int], _ k: Int) -> Int {
+        if nums.count == 0 {
+            return 0
+        }
+        var map: [Int: Int] = [:]
+        map[0] = 1 // 需要添加入一个元素垫底，已支持前几位就满足的情况
+        var presum = 0, count = 0
+
+        for x in nums {
+            presum += x
+            //当前前缀和已知，判断是否含有 presum - k的前缀和，那么我们就知道某一区间的和为 k 了。
+            if let v = map[presum - k] {
+                count += v //获取presum-k前缀和出现次数
+            }
+            map[presum] = (map[presum] ?? 0) + 1
+        }
+        return count
+    }
+}
+```

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

@@ -140,6 +140,43 @@ class Solution:
         return arr
 ```
 
+Swift Code:
+
+```swift
+class Solution {
+    func spiralOrder(_ matrix: [[Int]]) -> [Int] {
+        var arr:[Int] = []
+        var left = 0, right = matrix[0].count - 1
+        var top = 0, down = matrix.count - 1
+
+        while (true) {
+            for i in left...right {
+                arr.append(matrix[top][i])
+            }
+            top += 1
+            if top > down { break }
+            for i in top...down {
+                arr.append(matrix[i][right])
+            }
+            right -= 1
+            if left > right { break}
+            for i in stride(from: right, through: left, by: -1) {
+                arr.append(matrix[down][i])
+            }
+            down -= 1
+            if top > down { break}
+            for i in stride(from: down, through: top, by: -1) {
+                arr.append(matrix[i][left])
+            }
+            left += 1
+            if left > right { break}
+        }
+
+        return arr
+    }
+}
+```
+
 我们仅仅是将 54 反过来了，往螺旋矩阵里面插值，下面我们直接看代码吧,大家可以也可以对其改进，大家可以思考一下，如果修改能够让代码更简洁！
 
 Java Code:
@@ -226,3 +263,45 @@ class Solution:
         return arr.tolist()
 ```
 
+Swift Code:
+
+```swift
+class Solution {
+    func generateMatrix(_ n: Int) -> [[Int]] {
+        var arr:[[Int]] = Array.init(repeating: Array.init(repeating: 0, count: n), count: n)
+        var left = 0, right = n - 1
+        var top = 0, bottom = n - 1
+        var num = 1, numSize = n * n
+
+        while true {
+            for i in left...right {
+                arr[top][i] = num
+                num += 1
+            }
+            top += 1
+            if num > numSize { break}
+            for i in top...bottom {
+                arr[i][right] = num
+                num += 1
+            }
+            right -= 1
+            if num > numSize { break}
+            for i in stride(from: right, through: left, by: -1) {
+                arr[bottom][i] = num
+                num += 1
+            }
+            bottom -= 1
+            if num > numSize { break}
+            for i in stride(from: bottom, through: top, by: -1) {
+                arr[i][left] = num
+                num += 1
+            }
+            left += 1
+            if num > numSize { break}
+        }
+
+        return arr
+    }
+}
+```
+

animation-simulation/数组篇/leetcode66加一.md

@@ -85,3 +85,23 @@ class Solution:
         arr[0] = 1
         return arr
 ```
+
+Swift Code:
+
+```swift
+class Solution {
+    func plusOne(_ digits: [Int]) -> [Int] {
+        let count = digits.count
+        var digits = digits
+        for i in stride(from: count - 1, through: 0, by: -1) {
+            digits[i] = (digits[i] + 1) % 10
+            if digits[i] != 0 {
+                return digits
+            }
+        }
+        var arr: [Int] = Array.init(repeating: 0, count: count + 1)
+        arr[0] = 1
+        return arr
+    }
+}
+```

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

@@ -96,6 +96,38 @@ class Solution:
         nums[j] = temp
 ```
 
+Swift Code:
+
+```swift
+class Solution {
+    func sortColors(_ nums: inout [Int]) {
+
+        let count = nums.count
+        var left = 0, i = left, right = count - 1
+        while i <= right {
+            if nums[i] == 2 {
+                //nums.swapAt(i, right) 直接调用系统方法
+                self.swap(&nums, i, right) // 保持风格统一走自定义交换
+                right -= 1
+            } else if nums[i] == 0 {
+                //nums.swapAt(i, left) 直接调用系统方法
+                self.swap(&nums, i, left) // 保持风格统一走自定义交换
+                i += 1
+                left += 1
+            } else {
+                i += 1
+            }
+        }
+    }
+
+    func swap(_ nums: inout [Int], _ i: Int, _ j: Int) {
+        let temp = nums[i]
+        nums[i] = nums[j]
+        nums[j] = temp
+    }
+}
+```
+
 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]
@@ -174,4 +206,39 @@ class Solution:
         temp = nums[i]
         nums[i] = nums[j]
         nums[j] = temp
+```
+
+Swift Code:
+
+```swift
+class Solution {
+    func sortColors(_ nums: inout [Int]) {
+
+        let count = nums.count
+        var left = 0, i = left, right = count - 1
+        while i <= right {
+            if nums[i] == 0 {
+                //nums.swapAt(i, left) 直接调用系统方法
+                self.swap(&nums, i, left) // 保持风格统一走自定义交换
+                left += 1
+            } 
+            if nums[i] == 2 {
+                //nums.swapAt(i, right) 直接调用系统方法
+                self.swap(&nums, i, right) // 保持风格统一走自定义交换
+                right -= 1
+                //如果不等于 1 则需要继续判断，所以不移动 i 指针，i--
+                if nums[i] != 1 {
+                    i -= 1
+                }
+            }
+            i += 1
+        }
+    }
+
+    func swap(_ nums: inout [Int], _ i: Int, _ j: Int) {
+        let temp = nums[i]
+        nums[i] = nums[j]
+        nums[j] = temp
+    }
+}
 ```

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

@@ -61,6 +61,24 @@ class Solution:
         return -1
 ```
 
+Swift Code:
+
+```swift
+class Solution {
+    func findRepeatNumber(_ nums: [Int]) -> Int {
+        var set: Set<Int> = []
+        for n in nums {
+            if set.contains(n) { // 如果发现某元素存在，则返回
+                return n
+            }
+            set.insert(n) // 存入集合
+        }
+
+        return -1
+    }
+}
+```
+
 #### **原地置换**
 
 **解析**
@@ -136,3 +154,26 @@ class Solution:
                 nums[temp] = temp
         return -1
 ```
+
+Swift Code:
+
+```swift
+class Solution {
+    func findRepeatNumber(_ nums: [Int]) -> Int {
+        if nums.isEmpty {
+            return -1
+        }
+        var nums = nums;
+        for i in 0..<nums.count {
+            while nums[i] != i {
+                if nums[i] == nums[nums[i]] {
+                    return nums[i]
+                }
+                nums.swapAt(i, nums[i])
+            }
+        }
+
+        return -1
+    }
+}
+```

animation-simulation/数组篇/长度最小的子数组.md

@@ -106,3 +106,22 @@ class Solution:
             return windowlen
 ```
 
+Swift Code
+
+```swift
+class Solution {
+    func minSubArrayLen(_ target: Int, _ nums: [Int]) -> Int {
+
+        var sum = 0, windowlen = Int.max, i = 0
+        for j in 0..<nums.count {
+            sum += nums[j]
+            while sum >= target {
+                windowlen = min(windowlen, j - i + 1)
+                sum -= nums[i]
+                i += 1
+            }
+        }
+        return windowlen == Int.max ? 0 : windowlen
+    }
+}
+```