leetcode Weekly Contest 67 -- 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. Aswap 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

.

class Solution {
public:
int minSwapsCouples(vector<int>& row)
{
int a,sum=0;
for (int i = 0; i < row.size(); i += 2)
{

if (row[i] % 2 == 0)a = row[i] + 1;
else a = row[i] - 1;

for (int j = i + 1; j < row.size(); j++)
{
if (a == row[j]&& j != i + 1)
{
swap(row[i + 1], row[j]);
sum++;
break;
}
}
}

return sum;
}
};