[LeetCode] Sqrt(x)

 

Implement int sqrt(int x).

Compute and return the square root of x.

x is guaranteed to be a non-negative integer.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we want to return an integer, the decimal part will be truncated.

编写Sqrt(x)函数

使用二分搜索来查值进行平分计算。

class Solution {
public:
    int mySqrt(int x) {
        int left = 0, right = x;
        while (left <= right) {
            long long mid = left + (right - left) / 2;
            if (pow(mid, 2) > x) {
                right = mid - 1;
            }
            else {
                if (pow(mid + 1, 2) > x)
                    return mid;
                left = mid + 1;
            }
        }
    }
};
// 22 ms