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