leetcode69

public class Solution {
    public int MySqrt(int x) {
        long r = x;
            while (r * r > x)
                r = (r + x / r) / 2;
            return (int)r;
    }
}

https://leetcode.com/problems/sqrtx/#/description

补充一个python的实现：

class Solution:
    def mySqrt(self, x: int) -> int:
        r = x
        while r * r > x:
            r = (r + x // r) // 2
        return int(r)

 

使用内置函数：

class Solution:
    def mySqrt(self, x: int) -> int:
        return int(x ** 0.5)

 

Java的内置函数：

class Solution {
    public int mySqrt(int x) {
        double xx = Double.valueOf(x);
        return (int)Math.sqrt(xx);
    }
}