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.
题目标签：Bit Manipulation
这道题目给了我们两个int ， 让我们找出hamming distance。设一个for loop 循环32次，然后每一次循环，利用 & 1 来取得最后一个bit， 比较x和y是不是相等，不相等的话，增加res的值by 1。再利用 >> 1来移动bits向右一位。
Java Solution:
Runtime beats 57.75%
完成日期：06/28/2017
关键词：Bit Manipulation
关键点：利用 & 1拿到bit， 利用 >> 来移动bits
public class Solution
{
public int hammingDistance(int x, int y)
{
int res = 0; for(int i=0; i<32; i++)
{
if((x & 1) != (y & 1))
res++; x = x >> 1;
y = y >> 1;
} return res;
}
}
参考资料：N/A
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