leetcode 477. Total Hamming Distance

原题：

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.

代码如下：

int totalHammingDistance(int* nums, int numsSize) {    int total = 0;    int n = numsSize;    for (int j=0;j<32;j++)     {        int bitCount = 0;        for (int i=0;i<n;i++)        {            bitCount += (nums[i] >> j) & 1;        }        total += bitCount*(n - bitCount);    }    return total;}

思路如下，一开始想的是直接两两异或，然后直接统计1就好。想了下肯定会超时，所以就没写。。

想了半天没找到太好的规律。

就找了一下，

发现对32位int逐个进行统计1的个数，然后乘一下就能搞定。

这种思路还是很神作的。

PS：为何我睡觉早了反而觉得困了，，，