LeetCode(Others)Hamming Distance

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.

思路：利用转换二进制的函数，并将两个字符串相差的位置补0，最后统计有多少位不同的即可。

代码如下：

class Solution {
    public int hammingDistance(int x, int y) {
        String sx=Integer.toBinaryString(x);
        String sy=Integer.toBinaryString(y);
        int lenx=sx.length();
        int leny=sy.length();
        int lenmax=Math.max(lenx,leny);
        if(lenx<=leny){
            for(int i=0;i<leny-lenx;i++){
                sx="0"+sx;
            }
        }else{
            for(int i=0;i<lenx-leny;i++){
                sy="0"+sy;
            }
        }
        int count=0;
        for(int i=0;i<lenmax;i++){
            if(sx.charAt(i)!=sy.charAt(i)){
                count++;
            }
        }
        return count;
    }
}

　　还有一种按位亦或的算法，将两个输入的数字做按位异或，再统计得到的结果中1的个数就是这两个数的海明距离。

class Solution {
    public int hammingDistance(int x, int y) {
        int c=x^y;
        int result=0;
        for(int i=0;i<32;i++){
            if((c&1)==1){
                result++;
            }
            c=c>>>1;
        }
        return result;
    }
}