477. Total Hamming Distance(12/6/2017)

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：

4:   0100

14: 1110

2:   0010

加   2220 = 6

按列看，第一列一个1和两个0，相互的hamming distance为2；第二列一个0两个1，相互的hamming distance为2；第三列个0两个1，相互的hamming distance为2；第四列三个0，相互的hamming distance为0. 累加起来则为6

由此看出，我们只需要求得每一列的1的个数，即可知该列的hamming distance

代码：

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

出现过的小问题：+的优先级在&之上，所以((num >> i) & 1) + count此处应有两个括号。