leetcode-69. Sqrt(x)

題目：

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.

思路：

牛頓迭代法， 導數方程 f`(x)*x`=y`，  任何函式f(x)=y，求解某個y=n,均可以轉化為 f(x)-n=0,
        此後就可以用牛頓迭代法，不斷逼近實際待求x值。
        牛頓迭代共識：f`(x_pre)x_pre+x_pre=x_cur
應用: 迭代思想，類似於 動態規劃思想，pre==>cur，進行動態推斷處理

class Solution:
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        r=x/2+1
        while r*r-x>1e-10:
            r=(r+x/r)/2
            # print(r*r-x)