[LeetCode] 765. Couples Holding Hands 情侣牵手

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.

有N个情侣和2N个座位，想让每一对情侣都能够牵手，也就是挨着坐。每次能交换任意两个人的座位，求最少需要换多少次座位。

解法1：cyclic swapping

解法2: Union Find

Java: 1

public int minSwapsCouples(int[] row) {
    int res = 0, N = row.length;

    int[] ptn = new int[N];
    int[] pos = new int[N];

    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, j);
	    swap(pos, row[i], row[j]);
	    res++;
	}
    }

    return res;
}

private void swap(int[] arr, int i, int j) {
    int t = arr[i];
    arr[i] = arr[j];
    arr[j] = t;
}

Java: 2

class Solution {
    private class UF {
        private int[] parents;
        public int count;
        UF(int n) {
            parents = new int[n];
            for (int i = 0; i < n; i++) {
                parents[i] = i;
            }
            count = n;
        }

        private int find(int i) {
            if (parents[i] == i) {
                return i;
            }
            parents[i] = find(parents[i]);
            return parents[i];
        }

        public void union(int i, int j) {
            int a = find(i);
            int b = find(j);
            if (a != b) {
                parents[a] = b;
                count--;
            }
        }
    }
    public int minSwapsCouples(int[] row) {
        int N = row.length/ 2;
        UF uf = new UF(N);
        for (int i = 0; i < N; i++) {
            int a = row[2*i];
            int b = row[2*i + 1];
            uf.union(a/2, b/2);
        }
        return N - uf.count;
    }
}　

Python:

# Time:  O(n)
# Space: O(n)

class Solution(object):
    def minSwapsCouples(self, row):
        """
        :type row: List[int]
        :rtype: int
        """
        N = len(row)//2
        couples = [[] for _ in xrange(N)]
        for seat, num in enumerate(row):
            couples[num//2].append(seat//2)
        adj = [[] for _ in xrange(N)]
        for couch1, couch2 in couples:
            adj[couch1].append(couch2)
            adj[couch2].append(couch1)

        result = 0
        for couch in xrange(N):
            if not adj[couch]: continue
            couch1, couch2 = couch, adj[couch].pop()
            while couch2 != couch:
                result += 1
                adj[couch2].remove(couch1)
                couch1, couch2 = couch2, adj[couch2].pop()
        return result  # also equals to N - (# of cycles)　　

C++:

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;
}

　　

　　

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