477. Total Hamming Distance

标签：++   relevant   present   pairs   not   numbers   log   put   tco   

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:

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.

 1     public int totalHammingDistance(int[] nums) {
 2 //        仔细观察累计汉明距离和0跟1的个数，我们可以发现其实就是0的个数乘以1的个数，发现了这个重要的规律，
 3 //        那么整道题就迎刃而解了，只要统计出每一位的1的个数即可
 4         if (nums.length < 2) return 0;
 5         int result = 0;
 6         for (int i = 0; i < 32; i++) {
 7             int bitCount = 0;
 8             for (int number : nums) {
 9                 if ((number & (1 << i)) > 0) bitCount++;
10             }
11             result += bitCount * (nums.length - bitCount); //1的个数乘以0的个数
12         }
13         return result;  
14     }

477. Total Hamming Distance

标签：++   relevant   present   pairs   not   numbers   log   put   tco   

原文地址：http://www.cnblogs.com/wzj4858/p/7728186.html