题目
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.

答案

调用Hamming Distance这道题的答案O(n^2)次，不用想也知道会TLE，但我还是试了一下，结果却是是TLE。。

class Solution {
    public int numberOf1s(int x) {
        int count = 0;
        for(int i = 0; i < 32; i++) {
            int t = x >> i;
            if((t & 1) == 1) count++;
        }
        return count;
    }
    public int totalHammingDistance(int[] nums) {
        int total = 0;
        for(int i = 0; i < nums.length; i++) {
            for(int j = i + 1; j < nums.length; j++) {
                if(i != j) {
                    total += numberOf1s(nums[i]^nums[j]);   
                }
            }
        }
        return total;
    }
}

下面是一个O(n)的答案

    public int totalHammingDistance(int[] nums) {
        int sum = 0;
        for(int i = 0; i < 32; i++) {
            int zeros = 0;
            int ones = 0;
            for(int j = 0; j < nums.length; j++) {
                int t = nums[j] >> i;
                if((t & 1) == 1) ones++;
                else zeros++;
            }    
            sum += ones * zeros;
        }
        return sum;
    }