# 765. Couples Holding Hands 情侣牵手

## # 题目描述：

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.

## # 解题方法

[3 1 4 0 2 5]

[3 2 4 0 1 5]

[3 2 4 5 1 0]

``````class Solution(object):
def minSwapsCouples(self, row):
"""
:type row: List[int]
:rtype: int
"""
res = 0
n = len(row)
for i in range(0, n - 1, 2):
if row[i + 1] == (row[i] ^ 1):
continue
for j in range(i + 1, n):
if row[j] == (row[i] ^ 1):
row[j], row[i + 1] = row[i + 1], row[j]
res += 1
return res
``````

``````class Solution(object):
def minSwapsCouples(self, row):
"""
:type row: List[int]
:rtype: int
"""
n = len(row)
cnt = n / 2
self.par = range(n)
for i in range(0, n, 2):
x = self.find(row[i] / 2)
y = self.find(row[i + 1] / 2)
if x != y:
self.par[x] = y
cnt -= 1
return n / 2 - cnt

def find(self, x):
return x if self.par[x] == x else self.find(self.par[x])
``````

http://www.cnblogs.com/grandyang/p/8716597.html

## # 日期

2018 年 10 月 1 日 —— 欢度国庆！