[LeetCode] 477. Total Hamming Distance 全部汉明距离 [LeetCode] 191. Number of 1 Bits 二进制数1的个数 [LeetCode] 461. Hamming Distance 汉明距离 All LeetCode Questions List 题目汇总

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.

Python:

class Solution(object):
    def totalHammingDistance(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        result = 0
        for i in xrange(32):
            counts = [0] * 2
            for num in nums:
                counts[(num >> i) & 1] += 1
            result += counts[0] * counts[1]
        return result　　

C++:

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;
    }
};

　　

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[LeetCode] 191. Number of 1 Bits 二进制数1的个数

[LeetCode] 461. Hamming Distance 汉明距离

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