[Swift]LeetCode328. 奇偶鏈表 | Odd Even Linked List

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Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

Input: 
Output: 
1->2->3->4->5->NULL1->3->5->2->4->NULL

Example 2:

Input: 2
Output: 
->1->3->5->6->4->7->NULL2->3->6->7->1->5->4->NULL

Note:

• The relative order inside both the even and odd groups should remain as it was in the input.
• The first node is considered odd, the second node even and so on ...

---

給定一個單鏈表，把全部的奇數節點和偶數節點分別排在一塊兒。請注意，這裏的奇數節點和偶數節點指的是節點編號的奇偶性，而不是節點的值的奇偶性。

請嘗試使用原地算法完成。你的算法的空間複雜度應爲 O(1)，時間複雜度應爲 O(nodes)，nodes 爲節點總數。

示例 1:

輸入: 1->2->3->4->5->NULL
輸出: 1->3->5->2->4->NULL

示例 2:

輸入: 2->1->3->5->6->4->7->NULL 
輸出: 2->3->6->7->1->5->4->NULL

說明:

• 應當保持奇數節點和偶數節點的相對順序。
• 鏈表的第一個節點視爲奇數節點，第二個節點視爲偶數節點，以此類推。

---

36ms

 1 class Solution {
 2     func oddEvenList(_ head: ListNode?) -> ListNode? {
 3         let dummy = ListNode(0)
 4         var node = head
 5         var next = head?.next
 6         var oddEnd: ListNode?
 7         let evenHead = head?.next
 8         dummy.next = node
 9         var isOdd = true
10         while node != nil {
11             node?.next = node?.next?.next
12             if isOdd {
13                 oddEnd = node
14             }
15             node = next
16             next = next?.next
17             isOdd = !isOdd
18         }
19         oddEnd?.next = evenHead
20         return dummy.next
21     }
22 }

---

40ms

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public var val: Int
 5  *     public var next: ListNode?
 6  *     public init(_ val: Int) {
 7  *         self.val = val
 8  *         self.next = nil
 9  *     }
10  * }
11  */
12 class Solution {
13     func oddEvenList(_ head: ListNode?) -> ListNode? {
14         var odd: ListNode? = head
15         var even: ListNode? = head?.next
16         var evenFirst: ListNode? = even
17         
18         while true {
19             if odd == nil || even == nil || even?.next == nil {
20                 odd?.next = evenFirst
21                 break
22             }
23             
24             // Connecting odd
25             odd?.next = even?.next
26             odd = even?.next
27             
28             if odd?.next == nil {
29                 even?.next = nil
30                 odd?.next = evenFirst
31                 break
32             }
33             
34             // Connection even
35             even?.next = odd?.next
36             even = even?.next
37             
38         }
39         
40         return head
41     }
42 }

---

44ms

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public var val: Int
 5  *     public var next: ListNode?
 6  *     public init(_ val: Int) {
 7  *         self.val = val
 8  *         self.next = nil
 9  *     }
10  * }
11  */
12 class Solution {
13     func oddEvenList(_ head: ListNode?) -> ListNode? {
14         var oddNode = head?.next?.next
15         var evenNode = head?.next
16         let evenHeadNode = head?.next
17         var node = head
18         while oddNode != nil {
19             evenNode?.next = oddNode?.next
20             node?.next = oddNode
21             oddNode?.next = evenHeadNode
22             node = oddNode
23             evenNode = evenNode?.next
24             oddNode = evenNode?.next
25         }
26         
27         return head
28     }
29 }

---

56ms

 1 class Solution {
 2     func oddEvenList(_ head: ListNode?) -> ListNode? {
 3         let dummy = ListNode(0)
 4         let dummyOdd = ListNode(0)
 5         dummy.next = head
 6         var cur = head
 7         var curOdd = head?.next
 8         var index = 1
 9         dummyOdd.next = curOdd
10         while cur?.next != nil  && curOdd?.next != nil{
11             if index % 2 == 1 {
12                 cur?.next = curOdd?.next
13                 cur = cur?.next
14             }else {
15                 curOdd?.next = cur?.next
16                 curOdd = curOdd?.next
17             }
18             index += 1
19         }
20         cur?.next = dummyOdd.next
21         curOdd?.next = nil
22         return dummy.next
23     }
24 }

---

64ms

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public var val: Int
 5  *     public var next: ListNode?
 6  *     public init(_ val: Int) {
 7  *         self.val = val
 8  *         self.next = nil
 9  *     }
10  * }
11  */
12 class Solution {
13     func oddEvenList(_ head: ListNode?) -> ListNode? {
14         var i = head
15         var j = head?.next
16         let jH = j
17         
18         while i?.next?.next != nil || j?.next?.next != nil {
19             if i?.next?.next != nil {
20                 i?.next = i?.next?.next
21                 i = i?.next
22             }
23             if j?.next?.next != nil {
24                 j?.next = j?.next?.next
25                 j = j?.next
26             }
27         }
28         j?.next = nil // 清空偶數節點的next
29         i?.next = jH
30         return head
31     }
32 }

---

92ms

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public var val: Int
 5  *     public var next: ListNode?
 6  *     public init(_ val: Int) {
 7  *         self.val = val
 8  *         self.next = nil
 9  *     }
10  * }
11  */
12 class Solution {
13     func oddEvenList(_ head: ListNode?) -> ListNode? {
14         var first = head
15         var second = head?.next
16         let doulbeNode = second
17         var lastCNode = head
18         while first != nil || second != nil {
19             first?.next = second?.next
20             lastCNode = first
21             first = first?.next
22             second?.next = first?.next
23             second = second?.next
24         }
25         lastCNode?.next = doulbeNode
26         return head
27     }
28 }