LeetCode-Sqrt(x)

Description：
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 the decimal part is truncated, 2 is returned.

题意：模拟函数Sqrt(x)的功能，求一个数的开平方(取整数)；

解法：Java中函数sqrt(x)所使用的原理是“牛顿迭代法”

$$x_{n+1} = x_n - \frac{f(x_n)}{f^{'}(x_n)}$$

1. 对于我们要求的数r的平方根x，有等式$x^2=r$,即： $$f = x^2-r$$
2. 利用牛顿迭代法我们可以得到： $$x_{n+1} = x_n - \frac{f(x_n)}{f^{'}(x_n)}$$ $$x_{n+1} = x_n - \frac{x^2-r}{2x}$$ $$x_{n+1} = x_n - \frac{1}{2}(x-r/x)$$ $$x_{n+1} = \frac{1}{2}(x_n+r/x)$$
3. 在迭代的开始阶段，x的值可以取r，迭代的结束条件是$x^2 = r$，因此这里我们使用了一个误差$e^{-7}$

class Solution {
    public int mySqrt(int x) {
        double err = 1e-7;
        double result = x;
        while(Math.abs(result - x/result) > err){
            result = (result + x/result)/2.0;
        }
        return (int)result;
    }
}