LeetCode 477. Total Hamming Distance

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本文介绍了一种计算一组整数中所有数对间汉明距离之和的方法。通过位操作统计每位上1出现的次数，进而计算出每对数字间的汉明距离，最终求得总和。

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:

Elements of the given array are in the range of 0 to 10^9

Length of the array will not exceed 10^4.

class Solution {
public:
    int totalHammingDistance(vector<int>& nums) {
        int len = nums.size();
	    vector<int> v(numBit, 0);
	    for (auto digit : nums) {
		    int mask = 1;
		    for (int j = 0; j < numBit; ++j) {
			    v[j] += bool(digit & mask);
			    mask = mask << 1;
		    }
	    }
	    int result = 0;
	    for (int i = 0; i < numBit; ++i) {
		    result += v[i] * (len - v[i]);
	    }
	    return result;
    }
private:
    static const unsigned numBit = 32;
};

My github-leetcode: https://github.com/QiangL-DUT/LeetCode