Leetcode:Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

分析：二分查找。首先确定二分查找终止的条件和返回条件，其次对于与数字有关的题要注意int的表示范围防止溢出（比如该题两个int相乘可能会超过int的范围，故采用x/mid与mid比较大小）。代码如下：

class Solution {
public:
    int sqrt(int x) {
        if(x == 0) return 0;
        if(x == 1 || x == 2 || x == 3) return 1;
        int l = 1, r = x/2;
        
        while(l <= r){
            int mid = (l + r)/2;
            if(x/mid == mid || x/mid > mid && x/(mid+1) < (mid+1)) return mid;
            if(x/mid < mid) r = mid-1;
            else if(x/mid > mid) l = mid + 1;
        }
    }
};