TheHamming distancebetween 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 of0to10^9
2. Length of the array will not exceed10^4.

class Solution {
    public int totalHammingDistance(int[] nums) {
        int res = 0;
        int le = nums.length;
        for(int i = 0; i < 32; i++){
            int cur1 = 0;
            for
 
(int j = 0; j < nums.length; j++){
                    cur1 += ((nums[j]>>i) & 1);              
            }
            res += (cur1) * (le - cur1);
        }
        return res;
    }
}

經過觀察發現，每一位的漢明距離，等於該位1的個數*0的個數

然後就32位，每一位都統計1的數量然後做一次計算，返回