1 package BitManipulation; 2 //Question 461. Hamming Distance 3 /* 4 The Hamming distance between two integers is the number of positions at which the corresponding bits are different. 5 Given two integers x and y, calculate the Hamming distance. 6 Note: 7 0 ≤ x, y < 231. 8 Example: 9 Input: x = 1, y = 410 Output: 211 Explanation:12 1   (0 0 0 1)13 4   (0 1 0 0)14        ↑   ↑15 The above arrows point to positions where the corresponding bits are different.16  */17 public class hammingDistance461 {18     public static int hammingDistance(int x, int y) {19         int count=0;20         while(x>0|y>0){21             count+=(x&1)^(y&1);22             x=x>>1;23             y=y>>1;24         }25         return count;26     }27     //test28     public static void main(String[] args){29         int x=1;30         int y=4;31         System.out.println(hammingDistance(x,y));32     }33     34     //study the solution of other people35     //example1:use Integer.bitCount36     public static int hammingDistance1(int x, int y) {37         return Integer.bitCount(x ^ y);38     }39     //example2:xor = (xor) & (xor-1)40     //4(100),1(001)->101&100=100->100&011=00041     //1000001001这样中间的0可以一步直接跳过，机智！42     public int hammingDistance2(int x, int y) {43         int count = 0;44         for(int xor = x^y; xor != 0; xor = (xor) & (xor-1)) count++;45         return count;46     }47 }