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.

Approach #1: C++.

class Solution {
public:
    int minSwapsCouples(vector<int>& row) {
        int res = 0, N = row.size();
        
        vector<int> ptn(N, 0);
        vector<int> pos(N, 0);
        
        for (int i = 0; i < N; ++i) {
            ptn[i] = (i % 2 == 0 ? i+1 : i-1);
            pos[row[i]] = i;
        }
        
        for (int i = 0; i < N; ++i) {
            for (int j = ptn[pos[ptn[row[i]]]]; i != j; j = ptn[pos[ptn[row[i]]]]) {
                swap(row[i], row[j]);
                swap(pos[row[i]], pos[row[j]]);
                res++;
            }
        }
        
        return res;
    }
};

　　

Analysis:

https://leetcode.com/problems/couples-holding-hands/discuss/113362/JavaC%2B%2B-O(N)-solution-using-cyclic-swapping