[leetcode]765. Couples Holding Hands

链接：https://leetcode.com/problems/couples-holding-hands/description/

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.

思路：

对于row[0]来说，如果row[1] 和 row[0]是夫妻，那么他们两个就都不动就好。如果两个都移走，那么不是最优。 
考虑row[0]和row[1]不是夫妻的情况 
操作1: 要么把row[0]和 row[0]的夫妻边上的交换 
操作2: 要么把row[1]和row[0]的夫妻交换 
操作3：要么row[0]和row[1]都移走。 
显然操作3要不得。

对于0和1间隔奇数的情况: 
0abc1d 
变成 dabc10 或 01bcda 
对于0和1间隔偶数的情况 
0abcd1 
变成 01bcda 或者 dabc01

class Solution {
public:
    int minSwapsCouples(vector<int>& row) {
        int n = row.size();
        for (int i = 0; i < n; i++) {
            row[i] -= row[i] % 2;
        }
        int ans = 0;
        for (int i = 0; i < n; i += 2) {
            if (row[i] == row[i + 1]) {
                continue;
            }
            for (int j = i + 2; j < n; j++) {
                if (row[i] == row[j]) {
                    ans++;
                    swap(row[i + 1], row[j]);
                    break;
                }
            }
        }
        return ans;
    }
};