描述
The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
Given two integers x and y, return the Hamming distance between them.
Example 1:
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.
Example 2:
Input: x = 3, y = 1
Output: 1
Note:
0 <= x, y <= 2^31 - 1
解析
根据题意，就是计算汉明距离，两个整数之间的汉明距离是对应二进制位上不同数字的个数。先定义一个函数可以返回一个长度为 32 的表示数字十进制 n 的二进制的结果列表，然后找出变量 x 和 y 两个数字生成的两个列表中，对应索引上数字不相同的个数即可。
解答
class Solution(object):
def hammingDistance(self, x, y):
"""
:type x: int
:type y: int
:rtype: int
"""
def b(n):
r = [0]*32
i = 0
while n>0:
m = n%2
r[i] = m
n //= 2
i+=1
return r
x = b(x)
y = b(y)
result = 0
for i in range(32):
if x[i]!=y[i]:
result += 1
return result
运行结果
Runtime: 24 ms, faster than 13.74% of Python online submissions for Hamming Distance.
Memory Usage: 13.4 MB, less than 39.34% of Python online submissions for Hamming Distance.
解析
直接调用内置函数，找出 x 和 y 异或运算后的 1 的个数。
解答
class Solution(object):
def hammingDistance(self, x, y):
"""
:type x: int
:type y: int
:rtype: int
"""
return bin(x ^ y).count("1")
运行结果
Runtime: 12 ms, faster than 94.55% of Python online submissions for Hamming Distance.
Memory Usage: 13.5 MB, less than 39.34% of Python online submissions for Hamming Distance.
原题链接：leetcode.com/problems/ha…
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