【算法】——二分查找合集
阿华代码，不是逆风，就是我疯
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目录
零：二分查找工具
1：最基础模版
2：mid落点问题
一：最简单的二分
二：查找元素的位置（可能会有多个）
三：搜索插入位置
四：x的平方根
五：山脉数组的峰顶索引
六：寻找峰值
编辑
解法一
解法二
七：点名
零：二分查找工具
1：最基础模版
mid的写法可以防止溢出
2：mid落点问题
巧妙记忆：循环条件为while（left<right）时，left = mid，想象一下，只剩下两个球，那么我们的mid只能落在右端点，否则left = mid 会造成 left < right 死循环,此时我们确定的是右边的界限
重点：left 和right 根据题目的意思进行设置，然后才是mid的设置根据left和right的设置而设置（这才是这个二分查找的精髓所在）
简单记忆：落在哪个端点确定哪个界限
一：最简单的二分
704. 二分查找 - 力扣（LeetCode）
class Solution {public int search(int[] nums, int target) {//mid=left + (right - left)/3//用left移动思想来确定mid的位置，这种写法可以防溢出int left = 0 , right = nums.length-1 , mid = (left+right)/2;while(left<=right){if(nums[mid] < target){left = mid + 1 ;mid = (left+right)/2;}else if(nums[mid] > target){right = mid - 1;mid = (left+right)/2;}else{return mid;}}return -1;}
}二：查找元素的位置（可能会有多个）
34. 在排序数组中查找元素的第一个和最后一个位置 - 力扣（LeetCode）
class Solution {public int[] searchRange(int[] nums, int target) {int[] result = new int[]{-1,-1};if(nums.length == 0 ){return result;}//左端点int left = 0 , right = nums.length-1 ,targetLeft = 0 , targetRight = 0;while(left < right){int mid = left + (right - left )/2;if(nums[mid] < target){left = mid + 1;}else{right = mid;}}targetLeft = left;left = 0 ; right = nums.length-1 ;//右端点while(left < right){int mid = left + (right-left+1)/2;if(nums[mid] > target){right = mid - 1;}else{left = mid;}}targetRight = right;if(nums[targetLeft] != target){return result;}else if(nums[targetLeft] == nums[targetRight]){result[0] = targetLeft;result[1] = targetRight;return result;}else{}return result;}
}三：搜索插入位置
35. 搜索插入位置 - 力扣（LeetCode）
class Solution {public int searchInsert(int[] nums, int target) {if(nums.length == 0){return 0;}int targetLeft = 0 , n = nums.length;int left = 0 , right = nums.length-1;//这道题只用找一个左界限就够了//左界限left = 0 ; right = n-1;while(left < right){int mid = left + (right - left)/2;//左端点if(nums[mid] >= target){right = mid;}else{left = mid + 1;}} targetLeft = left;int result = 0;if(target > nums[targetLeft]){result = targetLeft + 1;}else{result = targetLeft ;}return result;}
}四：x的平方根
69. x 的平方根 - 力扣（LeetCode）
class Solution {public int mySqrt(int x) {long left = 0 , right = x ;if(x < 1 ){return 0;}long mid = 0;//mid的平方越界了while(left < right){mid = left + (right - left + 1)/2;if(mid * mid <= x){left = mid;}else{right = mid - 1 ;}}return (int)left;//强转为int类型}
}五：山脉数组的峰顶索引
852. 山脉数组的峰顶索引 - 力扣（LeetCode）
class Solution {public int peakIndexInMountainArray(int[] arr) {int left = 0 , right = arr.length , n = arr.length;while(left < right){int mid = left + (right - left + 1)/2;if(arr[mid] > arr[mid-1]){left = mid;}else if(arr[mid] < arr[mid-1]){right = mid - 1;}else{}}return left;}
}六：寻找峰值
162. 寻找峰值 - 力扣（LeetCode）
解法一
class Solution {public int findPeakElement(int[] nums) {int left = 0 , right = nums.length-1;//如果数组中只有一个元素，while循环都进不去，规避了这个问题nbwhile(left < right){int mid = left + (right - left )/2;if(nums[mid+1] > nums[mid]){left = mid + 1;}else if(nums[mid+1] < nums[mid]){right = mid;}else{}}return left;}
}解法二
class Solution {public int findPeakElement(int[] nums) {//暴力解法int n = nums.length , result = 0;if(n == 1){result = 0;}else if(nums[0] > nums[1]){result = 0;}else if(nums[n-1] > nums[n-2]){result = n-1;}else{int left = 0 , right = nums.length ;while(left < right){int mid = left + (right - left + 1)/2;if(nums[mid] > nums[mid-1]){left = mid;}else if(nums[mid] < nums[mid-1]){right = mid-1;}else{}}result = left;}return result;}
}七：寻找旋转排序数组中的最小值
153. 寻找旋转排序数组中的最小值
class Solution {public int findMin(int[] nums) {int left = 0 , n = nums.length , right = n-1;int tem = nums[n-1];while(left < right){int mid = left + (right - left)/2;if(nums[mid] <= nums[n-1]){right = mid;}else{left = mid + 1;}}return nums[left];}
}七：点名
LCR 173. 点名 - 力扣（LeetCode）
class Solution {public int takeAttendance(int[] records) {int left = 0 , n = records.length , right = records.length-1;if(records[0] != 0){return 0;}if(records[n-1] == n-1){return n;}while(left < right){int mid = left + (right - left)/2;if(records[mid] - mid <= 0){left = mid + 1;}else{right = mid ;}}return right;}
}