(二分查找 拓展) leetcode 69. 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

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 <= 1) return x;
        int left = 0;
        int right = x;
        while(left <= right){
            int mid = left + (right - left)/2;
            if(x / mid >= mid){
                left = mid + 1; 
            }
            else 
                right = mid - 1;
        }
        return right;
    }
};

 用sqrt：

class Solution {
public:
    int mySqrt(int x) {
        if(x <= 1)return x;
        return (int)(sqrt(x));
    }
};