package BitManipulation;
//Question 461. Hamming Distance
/*
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.
*/
public class hammingDistance461 {
public static int hammingDistance(int x, int y) {
int count=0;
while(x>0|y>0){
count+=(x&1)^(y&1);
x=x>>1;
y=y>>1;
}
return count;
}
//test
public static void main(String[] args){
int x=1;
int y=4;
System.out.println(hammingDistance(x,y));
} //study the solution of other people
//example1:use Integer.bitCount
public static int hammingDistance1(int x, int y) {
return Integer.bitCount(x ^ y);
}
//example2:xor = (xor) & (xor-1)
//4(100),1(001)->101&100=100->100&011=000
//1000001001这样中间的0可以一步直接跳过,机智!
public int hammingDistance2(int x, int y) {
int count = 0;
for(int xor = x^y; xor != 0; xor = (xor) & (xor-1)) count++;
return count;
}
}

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