477. Total Hamming Distance

477. 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, 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:

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.

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某一位上1的个数 ＊ 该位上0的个数 ＝ 该位上的hamming distance。遍历所有数字的32位，则得到结果

class Solution {
    public int totalHammingDistance(int[] nums) {
        int[] bits = new int[32];
        
        for (int num : nums) {
            
            int i = 0;
            while (num > 0) {
                if ((num & 0x1) == 1) {
                    bits[i]++;
                }
                num >>= 1;
                i++;
            }
        }
        
        int rt = 0;
        for (int bit : bits) {
            rt += bit * (nums.length - bit);
        }
        
        return rt;
    }
}