[Leetcode] 477. Total Hamming Distance 解题报告

题目：

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, 2Output: 6Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (justshowing 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.

思路：

1、暴力计算法：计算nums中每个pair中的Hamming距离，将和加起来即可。显然这种暴力法只能用来热身，过不了大数据测试。

2、按位统计法：注意到Hamming距离计算中，只有同一位置上的bit才会产生距离，所以我们可以统计nums中，32个位置上分别有多少个1和多少个0。假设在第i位有m个1和n个0，那么该位上贡献的Hamming距离就是m * n。最终将32个位上的贡献值加起来即可。

代码：

1、暴力统计法：

class Solution {public:    int totalHammingDistance(vector<int>& nums) {        int ret = 0;        for (int i = 0; i < nums.size(); ++i) {            for (int j = i + 1; j < nums.size(); ++j) {                ret += hammingDistance(nums[i], nums[j]);            }        }        return ret;    }private:    int hammingDistance(int a, int b) {        int c = a ^ b, ans = 0;        while (c != 0) {            ans += c & 1;            c >>= 1;        }        return ans;    }};

2、按位统计法：

class Solution {public:    int totalHammingDistance(vector<int>& nums) {        if (nums.size() == 0) {            return 0;        }        vector<int> zeros(32, 0), ones(32, 0);        for (auto num : nums) {            for (int i = 0; i < 32; ++i) {                if ((num >> i) & 1) {                    ++ones[i];                }                else {                    ++zeros[i];                }            }        }        int ans = 0;        for (int i = 0; i < 32; ++i) {            ans += zeros[i] * ones[i];        }        return ans;    }};