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
1 public class Solution 2 { 3 public int hammingDistance(int x, int y) 4 { 5 int res = 0; 6 7 for(int i=0; i<32; i++) 8 { 9 if((x & 1) != (y & 1)) 10 res++; 11 12 x = x >> 1; 13 y = y >> 1; 14 } 15 16 return res; 17 } 18 }
参考资料:N/A
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