[Swift]LeetCode765. 情侶牽手 | Couples Holding Hands

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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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.

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N 對情侶坐在連續排列的 2N 個座位上，想要牽到對方的手。 計算最少交換座位的次數，以便每對情侶能夠並肩坐在一塊兒。 一次交換可選擇任意兩人，讓他們站起來交換座位。

人和座位用 0 到 2N-1 的整數表示，情侶們按順序編號，第一對是 (0, 1)，第二對是 (2, 3)，以此類推，最後一對是 (2N-2, 2N-1)。

這些情侶的初始座位  row[i] 是由最初始坐在第 i 個座位上的人決定的。

示例 1:

輸入: row = [0, 2, 1, 3]
輸出: 1
解釋: 咱們只須要交換row[1]和row[2]的位置便可。

示例 2:

輸入: row = [3, 2, 0, 1]
輸出: 0
解釋: 無需交換座位，全部的情侶都已經能夠手牽手了。

說明:

1. len(row) 是偶數且數值在 [4, 60]範圍內。
2. 能夠保證row 是序列 0...len(row)-1 的一個全排列。

---

Runtime: 8 ms

Memory Usage: 19.1 MB

 1 class Solution {
 2     func minSwapsCouples(_ row: [Int]) -> Int {
 3         var row = row
 4         var res:Int = 0
 5         var n:Int = row.count
 6         for i in stride(from:0,to:n,by:2)
 7         {
 8             if row[i + 1] == (row[i] ^ 1)
 9             {
10                 continue            
11             }
12             res += 1
13             for j in (i + 1)..<n
14             {
15                 if row[j] == (row[i] ^ 1)
16                 {
17                     row[j] = row[i + 1]
18                     row[i + 1] = row[i] ^ 1
19                     break
20                 }
21             }
22         }
23         return res
24     }
25 }

---

8ms

 1 class Solution {
 2     func minSwapsCouples(_ row: [Int]) -> Int {
 3         guard row.count > 2 else {
 4             return 0 
 5         }
 6         var row = row 
 7         var numberOfSwaps: Int = 0 
 8         for seat in stride(from:0, to: row.count - 1, by: 2) {
 9             let current = row[seat]
10             let other = row[seat + 1]
11             if current == other ^ 1 { continue }
12             numberOfSwaps += 1 
13             for j in (seat+2)..<row.count {
14                 if row[j] == current ^ 1 {
15                     row.swapAt(j, seat + 1)
16                 }
17             }
18         }
19         return numberOfSwaps
20     }
21 }

---

12ms

 1 class Solution {
 2     func minSwapsCouples(_ row: [Int]) -> Int {
 3         var row = row
 4         var res = 0
 5         for i in 0..<row.count {
 6             guard i % 2 == 0 else { continue }
 7             let target = row[i] % 2 == 0 ? row[i] + 1 : row[i] - 1
 8             if row[i + 1] == target { continue }
 9             for j in i+2..<row.count {
10                 if row[j] == target {
11                     (row[i+1], row[j]) = (row[j], row[i+1])
12                     res += 1
13                 }
14             }
15         }
16         return res
17     }
18 }