Leetcode-Easy是Leecode难度为"Easy"的解法，由python编码实现。
461.Hamming Distance
- 描述：
- 思路：
首先将通过bin将x,y转化为二进制字符串，然后逆置，有利于后面字符串比较。用0补齐两个字符串，遍历字符串，比较对应的字符，得出结果。 - 代码
class Solution(object): def hammingDistance(self, x, y): """ :type x: int :type y: int :rtype: int """ result = 0 bx = bin(x)[2:][::-1] by = bin(y)[2:][::-1] if len(bx) <= len(by): bx, by = by, bx diff = len(bx) - len(by) for i in range(diff): by += '0' for i in range(len(bx)): if by[i] != bx[i]: result += 1 return result
另一种方法就是通过x和y的异或运算，统计1的个数。
def hammingDistance(self, x, y): """ :type x: int :type y: int :rtype: int """ return bin(x^y).count('1')
