[LeetCode] 69. Sqrt(x)_Easy tag: Binary Search

转载于 2018-08-14 23:53:00 发布 · 62 阅读

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原文链接：http://www.cnblogs.com/Johnsonxiong/p/9478896.html

本文介绍了一种使用牛顿法来计算整数平方根的方法。通过迭代计算，该算法可以高效地找到输入非负整数x的平方根的整数部分。文中提供了一个具体的Python实现示例。

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.

use Newton's method, the most important is ans = (ans + x/ans)/2

Code

class Solution:
    def sqrt(self, x):
        ans = x
        while ans * ans > x:
            ans = (ans + x//ans) //2
        return ans

转载于:https://www.cnblogs.com/Johnsonxiong/p/9478896.html