# 0477. Total Hamming Distance
###### tags: `Leetcode` `FaceBook` `Medium` `Bit Manipulation`
Link: https://leetcode.com/problems/total-hamming-distance/
## 思路
用count array记录所有数字的二进制表示里第i位是1的个数
ans = 对于所有i count[i] * (nums.length - count[i]) 求和
用位运算比不用位运算快很多
而且由于都是int数字，所以可以用31位的array存count number
## Code

class Solution {
    public int totalHammingDistance(int[] nums) {
        int[] count = new int[31];
        
        for(int num:nums){
            int power = 0;
            while(num > 0){
                count[power] += num & 1;
                num >>= 1;
                power++;
            }
        }
        
        int ans = 0;
        int size = nums.length;
        for(int cnt:count){
            ans += cnt*(size-cnt);
        }
        return ans;
    }
}