LeetCode 765. Couples Holding Hands

N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing any two people, then they stand up and switch seats.

The people and seats are represented by an integer from 0 to 2N-1, the couples are numbered in order, the first couple being (0, 1), the second couple being (2, 3), and so on with the last couple being (2N-2, 2N-1).

The couples’ initial seating is given by row[i] being the value of the person who is initially sitting in the i-th seat.

Example 1:

Input: row = [0, 2, 1, 3]
Output: 1
Explanation: We only need to swap the second (row[1]) and third (row[2]) person.

Example 2:

Input: row = [3, 2, 0, 1]
Output: 0
Explanation: All couples are already seated side by side.

Note:

1. len(row) is even and in the range of [4, 60].
2. row is guaranteed to be a permutation of 0...len(row)-1.

说是Hard，其实并不Hard。从前到后依次把应该配对的人配好对就行了。

class Solution {
public:
    int minSwapsCouples(vector<int>& row) {
        vector<int> seats(row.size());
        for (int i = 0; i < row.size(); i++) {
            seats[row[i]] = i;
        }
        
        int ans = 0;
        for (int i = 0; i < row.size(); i += 2) {
            int couple;
            if (row[i] % 2 == 0) {
                couple = row[i] + 1;
            }
            else {
                couple = row[i] - 1;
            }
            
            if (row[i + 1] != couple) {
                int coupleSeat = seats[couple];
                swap(seats[row[i + 1]], seats[couple]);
                swap(row[i + 1], row[coupleSeat]);
                ans++;
            }
        }
        return ans;
    }
};