# 477. 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:

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`.

4:     0 1 0 0

14:   1 1 1 0

2:     0 0 1 0

1:     0 0 0 1

``````class Solution {
public:
int totalHammingDistance(vector<int>& nums) {
int res = 0, n = nums.size();
for (int i = 0; i < 32; ++i) {
int cnt = 0;
for (int num : nums) {
if (num & (1 << i)) ++cnt;
}
res += cnt * (n - cnt);
}
return res;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/477

Hamming Distance

https://leetcode.com/problems/total-hamming-distance/

https://leetcode.com/problems/total-hamming-distance/discuss/96226/Java-O(n)-time-O(1)-Space

https://leetcode.com/problems/total-hamming-distance/discuss/96243/Share-my-O(n)-C%2B%2B-bitwise-solution-with-thinking-process-and-explanation

LeetCode All in One 题目讲解汇总(持续更新中…)

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