
class Solution {
    public int search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;  //右端点
//        int mid = (left + right)/2;   容易溢出
        while (left <= right){
            int mid = left + (right - left) / 2;  //先求区间的一半，防溢出
//            int mid = left + (right - left) / 3;  //区间的三分之一
            if (nums[mid] < target) left = mid+1;
            else if (nums[mid] > target) right = mid-1;
            else return mid;
        }
        return -1;
    }
}

2. 34.在排序数组中查找元素的第一个和最后一个位置 searchRange


class Solution {
    public int[] searchRange(int[] nums, int target) {
        //        处理边界
        int[] ret = new int[2];
        ret[0] = ret[1] = -1;
        if (nums.length == 0) return ret;
 
//        1，二分左端点
        int left = 0;
        int right = nums.length - 1;
        while (left<right){
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) left = mid +1;
            else right = mid;
        }
//        判断是否有结果
        if (nums[left] != target) return ret;
        else ret[0] = right;
 
//        2，二分右端点
        left = 0;
        right = nums.length-1;
        while (left<right){
            int mid = left + (right - left + 1) / 2;
            if (nums[mid] <= target) left = mid;
            else right = mid -1;
        }
        ret[1] = left;
        return ret;
    }
}

3. 35.搜索插入位置 searchInsert


class Solution {
    public int searchInsert(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left < right){
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) left = mid+1;
            else right = mid;
        }
        //target比最小的数小，放最左边
        if (nums[left] < target) return left + 1;
        return left;
    }
}

4. 69.×的平方根 mySqrt


class Solution {
    public int mySqrt(int x) {
        if(x < 1) return 0;
        long left = 1, right = x;
        while(left < right)
        {
            long mid = left + (right - left + 1) / 2;
            if(mid * mid <= x) left = mid;
            else right = mid - 1;
        }
        return (int)left;
    }
}

5. 852.山脉数组的峰顶索引 peakIndexInMountainArray


class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        int left = 1, right = arr.length - 2;
        while(left < right){
            int mid = left + (right - left + 1) / 2;
            if(arr[mid] > arr[mid - 1]) left = mid;
            else right = mid - 1;
        }
        return left;
    }
}

6. 162.寻找峰值 findPeakElement


class Solution {
    public int findPeakElement(int[] nums) {
        int left = 0;
        int right = nums.length - 1;
        while (left < right){
            int mid = left + (right - left) / 2;
            if (nums[mid] < nums[mid+1]) left = mid + 1;
            else right = mid;
        }
        return left;
    }
}

7. 153.寻找旋转排序数组中的最小值 findMin


class Solution {
    public int findMin(int[] nums) {
        int left = 0, right = nums.length - 1;
        int x = nums[right]; // 标记⼀下最后⼀个位置的值
        while(left < right){
            int mid = left + (right - left) / 2;
            if(nums[mid] > x) left = mid + 1;
            else right = mid;
        }
        return nums[left];
    }
}

8. LCR 173.点名 takeAttendance

一题多解，需要发散思维。

运用 哈希表，直接遍历，异或运算，高斯求和 都能求解出结果，时间复杂度都是O（n）。

但是运用二分查找来求解，时间复杂度就是O(logN)。


int left = 0;
        int right = records.length - 1;
        while (left < right){
            int mid = left + (right - left) / 2;
            if (records[mid] == mid) left = mid +1;
            else right = mid;
        }
//        细节eg，数组0,1,2,3，缺少4
        return records[left] == left ? left + 1 : left;

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