Open Shawngbk opened 7 years ago
就是位数不同的总数，所以先做^异或运算，再for循环，做移位运算，再做和1做&与 public class Solution { public int hammingDistance(int x, int y) { int xory = x ^ y, count = 0; for(int i = 0; i < 32; i++) { count += (xory >> i) & 1; } return count; } }
FB
Shawngbk
commented
7 years ago
就是位数不同的总数，所以先做^异或运算，再for循环，做移位运算，再做和1做&与 public class Solution { public int hammingDistance(int x, int y) { int xory = x ^ y, count = 0; for(int i = 0; i < 32; i++) { count += (xory >> i) & 1; } return count; } }