765. Couples Holding Hands

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.

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贪心，本质上Cycle Finding或union find
ref: https://leetcode.com/problems/couples-holding-hands/discuss/113362/JavaC++-O(N)-solution-using-cyclic-swapping

class Solution {
    public int minSwapsCouples(int[] row) {
        Map<Integer, Integer> valueToIndex = new HashMap<>();
        
        for (int i = 0; i < row.length; i++) {
            valueToIndex.put(row[i], i);
        }
        
        int swaps = 0;
        for (int i = 0; i < row.length; i += 2) {
            int p1 = row[i];
            int p2 = (p1 % 2) == 0 ? p1 + 1 : p1 - 1;
            
            if (row[i + 1] != p2) {
                int p2Index = valueToIndex.get(p2);
                valueToIndex.put(p2, i + 1);
                valueToIndex.put(row[i + 1], p2Index);
                swap(row, p2Index, i + 1);
                swaps++;
            }
        }
        
        return swaps;
    }
    
    private void swap(int[] row, int i1, int i2) {
        int tmp = row[i1];
        row[i1] = row[i2];
        row[i2] = tmp;
    }
}