leetcode461

    public class Solution
    {
        public int HammingDistance(int x, int y)
        {
            int[] aryA = new int[];
            int[] aryB = new int[];

            int i = ;
            int j = ;

            do
            {
                aryA[i] = x % ;//将10进制转换为2进制
                x = x / ;
                i++;
            }
            while (x != );

            do
            {
                aryB[j] = y % ;//将10进制转换为2进制
                y = y / ;
                j++;
            }
            while (y != );

            int result = ;
            for (int k = ; k < ; k++)
            {
                if (aryA[k] != aryB[k])//查找对应的二进制位，如果一个是0一个是1
                {
                    result++;
                }
            }
            //Console.WriteLine(result);
            return result;
        }
    }

https://leetcode.com/problems/hamming-distance/#/description

将10进制转为2进制

补充一个python的实现，使用位操作：

 class Solution:
     def hammingDistance(self, x: int, y: int) -> int:
         z = x ^ y
         cnt =
         mask =
         for i in range():
             t = z & mask
             mask <<=
             if t != :
                 cnt +=
         return cnt