477. Total Hamming Distance - Medium

标签：integer   length   div   AMM   each   amp   int   this   ace   

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.

For each bit position 1-32 in a 32-bit integer, we count the number of integers in the array which have that bit set. 

Then, if there are n integers in the array and k of them have a particular bit set and (n-k) do not, then that bit contributes k*(n-k) hamming distance to the total.

time = O(n), space = O(1)

class Solution {
    public int totalHammingDistance(int[] nums) {
        int res = 0, n = nums.length;
        for(int i = 0; i < 32; i++) {
            int oneCount = 0;
            for(int j = 0; j < nums.length; j++) {
                oneCount += (nums[j] >> i) & 1;
            }
            int zeroCount = n - oneCount;
            res += zeroCount * oneCount;
        }
        return res;
    }
}

477. Total Hamming Distance - Medium

标签：integer   length   div   AMM   each   amp   int   this   ace   

原文地址：https://www.cnblogs.com/fatttcat/p/11403392.html