[Swift]LeetCode765. 情侣牵手 | Couples Holding Hands

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： https://www.cnblogs.com/strengthen/p/10533329.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

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:

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.

---

N 对情侣坐在连续排列的 2N 个座位上，想要牵到对方的手。 计算最少交换座位的次数，以便每对情侣可以并肩坐在一起。 一次交换可选择任意两人，让他们站起来交换座位。

人和座位用 0 到 2N-1 的整数表示，情侣们按顺序编号，第一对是 (0, 1)，第二对是 (2, 3)，以此类推，最后一对是 (2N-2, 2N-1)。

这些情侣的初始座位  row[i] 是由最初始坐在第 i 个座位上的人决定的。

示例 1:

输入: row = [0, 2, 1, 3]
输出: 1
解释: 我们只需要交换row[1]和row[2]的位置即可。

示例 2:

输入: row = [3, 2, 0, 1]
输出: 0
解释: 无需交换座位，所有的情侣都已经可以手牵手了。

说明:

1. len(row) 是偶数且数值在 [4, 60]范围内。
2. 可以保证row 是序列 0...len(row)-1 的一个全排列。

---

Runtime: 8 ms

Memory Usage: 19.1 MB

class Solution {
    func minSwapsCouples(_ row: [Int]) -> Int {
        var row = row
        var res:Int = 0
        var n:Int = row.count
        for i in stride(from:0,to:n,by:2)
        {
            if row[i + 1] == (row[i] ^ 1)
            {
                continue            
            }
            res += 1
            for j in (i + 1)..<n
            {
                if row[j] == (row[i] ^ 1)
                {
                    row[j] = row[i + 1]
                    row[i + 1] = row[i] ^ 1
                    break
                }
            }
        }
        return res
    }
}

---

8ms

class Solution {
    func minSwapsCouples(_ row: [Int]) -> Int {
        guard row.count > 2 else {
            return 0 
        }
        var row = row 
        var numberOfSwaps: Int = 0 
        for seat in stride(from:0, to: row.count - 1, by: 2) {
            let current = row[seat]
            let other = row[seat + 1]
            if current == other ^ 1 { continue }
            numberOfSwaps += 1 
            for j in (seat+2)..<row.count {
                if row[j] == current ^ 1 {
                    row.swapAt(j, seat + 1)
                }
            }
        }
        return numberOfSwaps
    }
}

---

12ms

class Solution {
    func minSwapsCouples(_ row: [Int]) -> Int {
        var row = row
        var res = 0
        for i in 0..<row.count {
            guard i % 2 == 0 else { continue }
            let target = row[i] % 2 == 0 ? row[i] + 1 : row[i] - 1
            if row[i + 1] == target { continue }
            for j in i+2..<row.count {
                if row[j] == target {
                    (row[i+1], row[j]) = (row[j], row[i+1])
                    res += 1
                }
            }
        }
        return res
    }
}