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Merge pull request #42 from lin14543/main

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添加二分Go 代码
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算法基地 2021-07-29 11:46:29 +08:00 committed by GitHub
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8 changed files with 180 additions and 235 deletions
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34
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
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@ -40,6 +40,8 @@
#### **题目代码**
Java Code:
```java
class Solution {
public boolean search(int[] nums, int target) {
@ -73,35 +75,3 @@ class Solution {
}
}
```
Go Code:
```go
func search(nums []int, target int) bool {
l, r := 0, len(nums) - 1
for l <= r {
m := l + (r - l) / 2
if nums[m] == target {
return true
}
// 先判断哪边是递增的，再查找范围
if nums[m] == nums[l] {
l++
} else if nums[l] < nums[m] {
// 判断target是否在有序的那边就行了。
if nums[l] <= target && target < nums[m] {
r = m - 1
} else {
l = m + 1
}
} else if nums[l] > nums[m] {
if nums[m] < target && target <= nums[r] {
l = m + 1
} else {
r = m - 1
}
}
}
return false
}
```

20
animation-simulation/二分查找及其变种/leetcode153搜索旋转数组的最小值.md
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@ -104,23 +104,3 @@ public:
}
};
```
Go Code:
```go
func findMin(nums []int) int {
l, r := 0, len(nums) - 1
for l < r {
if nums[l] < nums[r] {
return nums[l]
}
m := l + (r - l) / 2
if nums[l] > nums[m] {
r = m
} else {
l = m + 1
}
}
return nums[l]
}
```

34
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
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@ -89,6 +89,8 @@
#### 题目代码
Java Code:
```java
class Solution {
public int search(int[] nums, int target) {
@ -126,35 +128,3 @@ class Solution {
}
}
```
Go Code:
```go
func search(nums []int, target int) int {
l, r := 0, len(nums) - 1
for l <= r {
m := (l + r) / 2
if target == nums[m] {
return m
}
// 先判断哪边是有序的
if nums[m] < nums[r] {
// 再判断target在左右哪边
if target > nums[m] && target <= nums[r] {
l = m + 1
} else {
r = m - 1
}
} else {
if target < nums[m] && target >= nums[l] {
r = m - 1
} else {
l = m + 1
}
}
}
return -1
}
```
##

44
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
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@ -122,6 +122,8 @@ int upperBound(int[] nums, int target) {
#### **题目完整代码**
Java Code:
```java
class Solution {
public int[] searchRange (int[] nums, int target) {
@ -166,45 +168,3 @@ class Solution {
}
}
```
Go Code:
```go
func searchRange(nums []int, target int) []int {
upper := upperBound(nums, target)
lower := lowerBound(nums, target)
if (upper < lower) {
return []int{-1, -1}
}
return []int{lower, upper}
}
// upperBound 计算上边界
func upperBound(nums []int, target int) int {
l, r := 0, len(nums) - 1
for l <= r {
m := l + (r - l) / 2
if target >= nums[m] {
l = m + 1
} else if target < nums[m] {
r = m - 1
}
}
return r
}
// lowerBound 计算下边界
func lowerBound(nums []int, target int) int {
l, r := 0, len(nums) - 1
for l <= r {
m := l + (r - l) / 2
if target <= nums[m] {
r = m - 1
} else if target > nums[m] {
l = m + 1
}
}
return l
}
```

20
animation-simulation/二分查找及其变种/leetcode35搜索插入位置.md
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@ -40,6 +40,8 @@
#### 题目代码
Java Code:
```java
class Solution {
public int searchInsert(int[] nums, int target) {
@ -67,21 +69,3 @@ class Solution {
```
Go Code:
```go
func searchInsert(nums []int, target int) int {
l, r := 0, len(nums) - 1
for l <= r {
m := l + (r - l) / 2
if target == nums[m] {
return m
}
if target < nums[m] {
r = m - 1
} else if target > nums[m] {
l = m + 1
}
}
return l
}
```

54
animation-simulation/二分查找及其变种/二分查找详解.md
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@ -94,8 +94,10 @@ target < nums[mid] 则在左半区间继续进行搜索，即 right = mid -1；
下面我们来看一下二分查找的代码，可以认真思考一下 if 语句的条件，每个都没有简写。
Java Code:
```java
public static int binarySearch(int[] nums,int target,int left, int right) {
public static int binarySearch(int[] nums,int target,int left, int right) {
//这里需要注意，循环条件
while (left <= right) {
//这里需要注意，计算mid
@ -112,7 +114,30 @@ target < nums[mid] 则在左半区间继续进行搜索，即 right = mid -1；
}
//没有找到该元素，返回 -1
return -1;
}
```
Go Code:
```go
func binarySearch(nums []int, target, left, right int) int {
//这里需要注意，循环条件
for left <= right {
//这里需要注意，计算mid
mid := left + ((right - left) >> 1)
if nums[mid] == target {
return mid
} else if nums[mid] < target {
//这里需要注意，移动左指针
left = mid + 1
} else if nums[mid] > target {
//这里需要注意，移动右指针
right = mid - 1
}
}
//没有找到该元素，返回 -1
return -1
}
```
二分查找的思路及代码已经理解了，那么我们来看一下实现时容易出错的地方
@ -127,6 +152,8 @@ target < nums[mid] 则在左半区间继续进行搜索，即 right = mid -1；
下面我们来看一下二分查找的递归写法
Java Code:
```java
public static int binarySearch(int[] nums,int target,int left, int right) {
@ -145,5 +172,28 @@ public static int binarySearch(int[] nums,int target,int left, int right) {
}
//不存在返回-1
return -1;
}
}
```
Go Code:
```go
func binarySearch(nums []int, target, left, right int) int {
if left <= right {
mid := left + ((right - left) >> 1)
if nums[mid] == target {
//查找成功
return mid
} else if nums[mid] > target {
//新的区间,左半区间
return binarySearch(nums, target, left, mid-1)
} else if nums[mid] < target {
//新的区间，右半区间
return binarySearch(nums, target, mid+1, right)
}
}
//不存在返回-1
return -1
}
```

29
animation-simulation/二分查找及其变种/二维数组的二分查找.md
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@ -52,6 +52,8 @@
#### 题目代码
Java Code:
```java
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
@ -83,30 +85,3 @@ class Solution {
}
}
```
Go Code:
```go
func searchMatrix(matrix [][]int, target int) bool {
if len(matrix) == 0 {
return false
}
row, col := len(matrix), len(matrix[0])
// 将二维数组拉平为一维。
l, r := 0, row * col - 1
for l <= r {
m := l + (r - l) / 2
// 将一维的坐标转换为二维
x, y := m / col, m % col
if matrix[x][y] == target {
return true
} else if matrix[x][y] < target {
l = m + 1
} else if matrix[x][y] > target {
// 二分查找时，把所有的情况都要写出来，避免直接else
r = m - 1
}
}
return false
}
```

58
animation-simulation/二分查找及其变种/找出第一个大于或小于目标的索引.md
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@ -34,6 +34,8 @@
OK!我们到这一步就能把这个变种给整的明明白白的了，下面我们看一哈程序代码吧，也是非常简单的。
Java Code:
```java
public static int lowBoundnum(int[] nums,int target,int left, int right) {
@ -56,7 +58,33 @@ public static int lowBoundnum(int[] nums,int target,int left, int right) {
}
//所有元素都小于目标元素
return -1;
}
```
Go Code:
```go
func lowBoundnum(nums []int, target, left, right int) int {
for (left <= right) {
//求中间值
mid := left + ((right - left) >> 1);
//大于目标值的情况
if (nums[mid] > target) {
//返回 mid
if (mid == 0 || nums[mid-1] <= target) {
return mid
}else{
right = mid -1
}
} else if (nums[mid] <= target){
left = mid + 1
}
}
//所有元素都小于目标元素
return -1
}
```
## **找出最后一个小于目标元素的索引**
@ -67,6 +95,8 @@ public static int lowBoundnum(int[] nums,int target,int left, int right) {
查找最后一个小于目标数的元素，比如我们的目标数为 7 ，此时他前面的数为 6，最后一个 6 的索引为 5，此时我们返回 5 即可，如果目标数元素为 12，那么我们最后一个元素为 11，仍小于目标数，那么我们此时返回 8，即可。这个变种其实算是上面变种的相反情况，上面的会了，这个也完全可以搞定了，下面我们看一下代码吧。
Java Code:
```java
public static int upperBoundnum(int[] nums,int target,int left, int right) {
@ -89,5 +119,31 @@ public static int upperBoundnum(int[] nums,int target,int left, int right) {
}
//没有查询到的情况
return -1;
}
}
```
Go Code:
```go
func upperBoundnum(nums []int, target, left, right int) int {
for left <= right {
mid := left + ((right - left) >> 1)
//小于目标值
if nums[mid] < target {
//看看是不是当前区间的最后一位，如果当前小于，后面一位大于，返回当前值即可
if mid == right || nums[mid+1] >= target {
return mid
} else {
left = mid + 1
}
} else if nums[mid] >= target {
right = mid - 1
}
}
//没有查询到的情况
return -1
}
```