LeetCode 765. Couples Holding Hands

最新推荐文章于 2025-06-19 07:15:00 发布

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本文探讨了一个有趣的问题，即如何通过最少的座位交换让每对情侣坐在一起。利用贪心算法，作者详细解释了解决方案，并提供了一段Java代码实现。该算法将情侣之间的座位调整视为环状结构，通过计算环的数量来确定最少交换次数。

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 swapconsists 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:

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

看到提示里有贪心算法，就确定了思考方向。我的思路是，一系列互相纠缠的情侣构成了一个环，而n对情侣构成的环至少要n-1次交换才能调整好。例如说[0,31,......,30,9,........,8,1,.......]这三对情侣就构成了一个环，只需要调整两次就可以让这三对情侣牵上手。

不同环之间的情侣不会互相干涉，因此也不存在不同环之间互相换使调整次数更少的情况了。

下面是我的代码：

public int minSwapsCouples(int[] row) {
        int result = 0;
        int[] sit = new int[row.length];
        for(int i=0;i<row.length;i++)
        {
        	sit[row[i]]=i;
        }
        int[] record = new int[row.length/2];
        int target,pos;
        for(int i = 0;i<row.length;i+=2)
        {
        	if(record[i/2]==1)
        		continue;
        	target = Partner(row[i]);
        	pos = Partner(i);
        	result++;
        	while(row[Partner(sit[Partner(row[pos])])]!=target)
        	{
        		pos = Partner(sit[Partner(row[pos])]);
                record[pos/2]=1;
        		result++;
        	}
        	record[sit[Partner(row[pos])]/2]=1;
        }
        return result;
    }
	
	public int Partner(int i)
	{
		if(i%2==0)
			return i+1;
		else return i-1;
	}