Total Hamming Distance

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

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, 2

Output: 6

Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just

showing 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:

Elements of the given array are in the range of 0 to 10^9

Length of the array will not exceed 10^4.

Thoughts

一对整数的海明距离定义为它们二进制不一样的位数，给定一组整数，问他们俩俩之间海明距离的总和。根据例子可观察到对每一位，每个1都可以和其它的0形成一个海明距离。因此对整数的32位每位都统计有多少1，count_1 * (N - count_1)就是当前位所增加的海明距离。

Code

/*
 * @lc app=leetcode id=477 lang=cpp
 *
 * [477] Total Hamming Distance
 */

// @lc code=start
class Solution {
public:
    int totalHammingDistance(vector<int>& nums) {
        const int N = nums.size();
        int res = 0;
        for (int i = 0; i < 32; ++i) {
            int one_c = 0;
            for (auto num : nums) {
                if ((num >> i) & 1 == 1) ++one_c; 
            }
            res += one_c * (N - one_c);
        }
        return res;
    }
};
// @lc code=end

class Solution {
    public int totalHammingDistance(int[] nums) {
        int res = 0;
        for (int i = 0; i < 32; i++) {
            int oneCount = 0;
            for (int num : nums) {
                oneCount += (num >> i) & 1;
            }
            res += oneCount * (nums.length - oneCount);
        }
        return res;
    }
}

Analysis

时间O(N).