[leetcode]Sqrt(x)

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: 4
Output:2

分析：

求非负数的平方根，可以用二分法来求解。首先x的平方根肯定是小于x/2+1的，所以i和j分别从0和x/2+1开始，计算（i+j）/2的平方，若比x小，则i+1，若比x大，则j-1，直到相等为止。

class Solution {
public:
    int mySqrt(int x) {
        //二分法求解
        long i = 0;
        long j = x/2+1;//x的平方根<=x/2+1
        while(i <= j)
        {
            long mid = (i+j)/2;
            long s = mid*mid;
            if(s == x)
                return mid;
            else if(s < x)
                i = mid+1;
            else
                j = mid-1;                                           
        }
        return j;
        
    }
};