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

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

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

原文：https://www.cnblogs.com/Johnsonxiong/p/9478896.html