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

Example 2:

Input: 8

Output: 2

Explanation: The square root of 8 is 2.82842..., and since

             the decimal part is truncated, 2 is returned.
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这个问题，其实很简单（用sqrt直接OK了），不过，我用二分查找解决这个题
C++代码：

class Solution {

public:

    int mySqrt(int x) {

        if(x <= ) return x;

        int left = ;

        int right = x;

        while(left <= right){

            int mid = left + (right - left)/;

            if(x / mid >= mid){

                left = mid + ;

            }

            else

                right = mid - ;

        }

        return right;

    }

};

用sqrt：

class Solution {

public:

    int mySqrt(int x) {

        if(x <= )return x;

        return (int)(sqrt(x));

    }

};