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:
len(row) is even and in the range of [4, 60].
row is guaranteed to be a permutation of 0…len(row)-1.
思路：
贪心，不过不好证明。从位置0开始，看位置1是否是它的partner，如果是，直接忽略这组couple，如果不是，从后面的数组中，找到partner直接交换，并且记录交换次数一次。遍历完整个数组即为最优解。
简单举例说明下：
首先，如果相邻两个位置已经是couple了，则直接忽略，那么原问题就变成了两种情况：
1. 交换位置后，能够使得两组couple得到匹配（最优）
2. 交换位置后，只能使得一组couple得到匹配（次优）
显然我们会进行最优的couple匹配，先把这部分给匹配了。针对上述贪心的做法，的确可以解决问题1.
问题2的关键在于不同的匹配顺序，是否会影响到子问题的最优解？答案是否定的。比如
[a, c, b, e, d, f]
可以得到三组block如下： a — c b — e d — f
考虑a — c 的子状态：
1. 匹配a — b,得到 c — e d — f
2. 匹配c — d,得到 b — e a — f
两种可能的子状态实际是等价的，所以无关乎顺序，直接贪心选择a的partner或者c的partner即可。
class Solution {
public int minSwapsCouples(int[] row) {
int n = row.length;
int ans = 0;
for (int i = 0; i < n; i += 2) {
int dest;
if ((row[i] & 1) == 0) dest = row[i] + 1;
else dest = row[i] - 1;
if (row[i + 1] == dest) continue;
ans ++;
for (int j = i + 2; j < n; ++j) {
if (row[j] == dest) {
int tmp = row[i + 1];
row[i + 1] = row[j];
row[j] = tmp;
break;
}
}
}
return
