461.Hamming Distance --leetcode

很久没有编程了, 想锻炼锻炼自己的编程能力, 准备开始刷些编程题, 以前学得那点C已经忘得差不多了, 所以准备用python编, 之后可能会有C++或者Java的编程解法, 不过其实leetcode上有各种语言的解法, 很多都写的好棒…

题目

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

思路

题目说的很清楚了,计算汉明距离的话, 对两个字符串进行异或运算，并统计结果为1的个数,这就是汉明距离

python解法

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class Solution(object):
    def hammingDistance(self, x, y):
        """
        :type x: int
        :type y: int
        :rtype: int
        """
        return bin(x^y).count('1')