leetcode15 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.

给一个非负数x整形，然后给其开方，得到结果，如果结果是一个非整形，需要只取整数部分。

我直接就用了java中的Math类中的sqrt函数，然后对其调用结果，进行了强制类型转化，去掉小数点后面的值。

代码实现如下：

class Solution {
    public int mySqrt(int x) {
        return (int)Math.sqrt(x);
    }
}

Runtime: 1 ms, faster than 100.00% of Java online submissions for Sqrt(x).
Memory Usage: 32.2 MB, less than 100.00% of Java online submissions for Sqrt(x).

然而我想看一下java中sqrt的源码，但是它是StrictMath类中的一个静态本地方法，具体的实现没有办法看到。

discuss中的一些解法：

关于sqrt(x)的数学思想：https://en.wikipedia.org/wiki/Integer_square_root#Using_only_integer_division

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

        return (int) r;
    }
}

class Solution {
    public int mySqrt(int x) {
         if(x <= 0) return 0;
        int l = 1, r = x, res = 1;
        while(l < r) {
            int mid = (l + r) / 2;
            if(mid > x/mid) {
                r = mid;
            } else {
                res = mid;
                l = mid + 1;
            }
        }
        return res;
    }
}