二分法

leetcode-34:寻找左右边界
Tips：
对于闭区间来讲，l <= r 是比较便于理解的

如何左右边界分别取左右哪个指针判断？假设指针相遇的时候，我需要得到左边界，有两种得到的方法，一是让左指针倒反天罡往左走格，二是要右指针向左走一格，同时要满足跳出while(l<=r)的循环，必须要满足 l > r,那么就必然是右指针执行最后一动跳出循环

如何理解得到左右边界后的条件判断？假设target在左边界外，那么一直在动的一定是它的右指针，右指针对应的左边界，那么左边界一定是不断更新的，右指针一定是不更新的，所以右指针为-2，同理在右边界以外的数字左边界是-2，便可以判断

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int leftborder = getLeftBorder(nums,target);
        int rightborder = getRightBorder(nums,target);
        if(leftborder == -2 || rightborder == -2)return {-1,-1};
        if ( rightborder - leftborder > 1)return {leftborder+1,rightborder-1};
        return {-1,-1};
    }
private:
    int getLeftBorder(vector<int>& nums,int target){
        int left = 0 , right = nums.size() - 1 , middle , leftborder = -2;
        while(left <= right){
            middle = left + (right - left) / 2;
            if(nums[middle] >= target){
                right = middle - 1;
                leftborder = right;
            }else{
                left = middle + 1;
            }
        }
        return leftborder;
    }
private:
    int getRightBorder(vector<int>& nums,int target){
        int left = 0 , right = nums.size() - 1 , middle , rightborder = -2;
        while(left <= right){
            middle = left + (right - left) / 2;
            if(nums[middle] > target){
                right = middle - 1;
            
            }else{
                left = middle + 1;
                rightborder = left;
            }
        }
        return rightborder;
    }
};

leetcode-69:平方根
Tips:
每次写完以后代入数据试一下，注意return的是right不是middle

为什么不return left？这样想 1的平方根才是他自己，倘若小于1，其平方根才可能要return左边界

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