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

2018-08-14 23:53  Johnson_强生仔仔  阅读(167)  评论(0编辑  收藏  举报

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