LeetCode: 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:

Elements of the given array are in the range of 

0 

to 

10^9

Length of the array will not exceed 

10^4

.

思路：一个有多个0,1组成的序列，所有的汉明距离和为0的个数乘以1的个数

int totalHammingDistance(int* nums, int numsSize) {

int totalD = 0;
int bits[32];
for (int i = 0; i < 32; ++i) {
bits[i] = 0;
}

for (int i = 0; i < numsSize; ++i) {
for (int j = 0; j < 32; ++j) {
bits[j] += nums[i] & 1;
nums[i] >>= 1;
}
}

for (int i = 0; i < 32; ++i) {
totalD += bits[i] * (numsSize - bits[i]);
}

return totalD;
}