477. Total Hamming Distance(C++)

原題

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.

解題思路

一開始我是寫了算法，先兩兩數字異或，然後再把為1的數字加起來，再把所有配對的結果加起來。之後我參考了別人的寫法，直接把所有數字的同一位加起來，這樣在時間上會節省很多。

代碼

class Solution {
public:
    int totalHammingDistance(vector<int>& nums) {
        int length = nums.size(), bit[32] = {0}, result = 0;
        
        for (int i = 0; i < 32; i++) {
            for (int j = 0; j < length; j++) {
                bit[i] += (nums[j]>>i)&1;
            }
            result += (length-bit[i])*bit[i];
        }
        return result;
    }
};

感想

這題不是很難，懂二進制的話一看題目就清楚要咋做了。但是一開始思路就往題目說的那種方法偏了，兩兩配對面對大量的數字會浪費极大的時間，遠不如直接按位算的速度。