二分搜索常用【转载】

转自：https://www.cnblogs.com/grandyang/p/6854825.html

1.查找和目标值完全相等的数

int find(vector<int>& nums, int target) {
    int left = 0, right = nums.size();
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) return mid;
        else if (nums[mid] < target) left = mid + 1;
        else right = mid;
    }
    return -1;
}

  这里给出了一个总结，关于right、循环条件、return的这些控制。

 

对于我来说的话，我还是比较喜欢，第二套？比较习惯，right=len(nums)-1这样，那么之后right=mid-1，return也是right-1，while循环的话需要等号。

2.寻找左边界

转自：https://labuladong.gitee.io/algo/2/18/28/

// 搜索左侧边界
int left_bound(int[] nums, int target) {
    if (nums.length == 0) return -1;
    int left = 0, right = nums.length;
    
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            // 当找到 target 时，收缩右侧边界
            right = mid;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid;
        }
    }
    return left;
}

3.寻找右边界

// 搜索右侧边界
int right_bound(int[] nums, int target) {
    if (nums.length == 0) return -1;
    int left = 0, right = nums.length;
    
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            // 当找到 target 时，收缩左侧边界
            left = mid + 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid;
        }
    }
    return left - 1;
}