Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4Output: 2

Example 2:

Input: 8Output: 2Explanation: The square root of 8 is 2.82842..., and since              the decimal part is truncated, 2 is returned.

 

• 用循环实现二分法
• 用除法防止溢出 

class Solution {    public int mySqrt(int x) {        int left = 1;        int right = (x>>1) + 1;        int mid = 1;        while(left <= right){            mid = left + ((right-left) >> 1);            if(mid < x/mid){ //用除法防止溢出                left = mid+1;            }            else if(mid > x/mid){                right = mid-1;            }            else return mid;        }        return right;    }    }