LeetCode 477. Total Hamming Distance

原题网址：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.

方法一：穷举搜索，时间复杂度O(N^2)，无法通过测试。

方法二：按照比特分组统计。

public class Solution {    public int totalHammingDistance(int[] nums) {        int total = 0;        for(int i = 0, bit = 1; i < 31; i++, bit <<= 1) {            int zeros = 0, ones = 0;            for(int num : nums) {                if ((num & bit) == 0) zeros++;                else ones++;            }            total += zeros * ones;        }        return total;    }}