[leetcode]477. Total Hamming Distance

题目链接:https://leetcode.com/problems/total-hamming-distance/

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:

Input: 4, 14, 2Output: 6Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (justshowing the four bits relevant in this case). So the answer will be:HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

Note:

1. Elements of the given array are in the range of 0 to 10^9
2. Length of the array will not exceed 10^4.

解题思路：数组大小为size，统计每个数在同一个bit位的出现1的个数，记为bitCount,那么这个bit位贡献的汉明距离为bitCount*(size-bitCount)。总的时间复杂度为O(size)。

class Solution{public:    int totalHammingDistance(vector<int>& nums)    {        int count=0;        for(int i=0;i<32;i++)        {            int bitCount=0;            for(int j=0;j<nums.size();j++)            {                bitCount+=(nums[j]>>i)&1;            }            count+=bitCount*(nums.size()-bitCount);        }        return count;    }};