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.
class Solution { public int hammingDistance(int x, int y) { String xx = Integer.toBinaryString(x); String yy = Integer.toBinaryString(y); int res = 0; System.out.println(xx+" " + yy); if(xx.length() > yy.length()){ int d = xx.length() - yy.length(); while(d > 0){ yy = "0" + yy; d--; } } if(yy.length() > xx.length()){ int d = yy.length() - xx.length(); while(d > 0){ xx = "0" + xx; d--; } } for(int i = 0; i < xx.length(); i++){ if(xx.charAt(i) != yy.charAt(i)) res++; } return res; } }
不够的前面补0
class Solution { public int hammingDistance(int x, int y) { return Integer.bitCount(x^y); } }
大佬的做法。。这就是差距吧，，比不同用异或，有多少位不同就有多少个1，然后bitCount完事
原文：https://www.cnblogs.com/wentiliangkaihua/p/12886570.html