[Swift]LeetCode206. 反轉鏈表 | Reverse Linked List

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➤微信公衆號:山青詠芝(shanqingyongzhi)
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Reverse a singly linked list.node

Example:git

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

Follow up:github

A linked list can be reversed either iteratively or recursively. Could you implement both?微信


反轉一個單鏈表。app

示例:ide

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

進階:
你能夠迭代或遞歸地反轉鏈表。你可否用兩種方法解決這道題?spa


20mscode

 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 reverseList(_ head: ListNode?) -> ListNode? {
14         if head == nil || head?.next == nil
15         {
16             return head
17         }
18         var h = reverseList(head?.next)
19         head?.next?.next = head
20         head?.next = nil
21         return h
22     }
23 }

20mshtm

 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 reverseList(_ head: ListNode?) -> ListNode? {
14         if head == nil {
15             return nil
16         }
17         
18         var pre : ListNode?
19         var next : ListNode?
20         var cur : ListNode? = head
21 
22         while (cur != nil) {
23             next = cur?.next;
24             cur?.next = pre;
25             pre = cur;
26             cur = next;
27         }
28         
29         return pre;
30     }
31 }

20ms

 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 reverseList(_ head: ListNode?) -> ListNode? {
14         if head == nil {
15             return head
16         }
17         
18         if head?.next == nil {
19             return head
20         }
21         
22         guard let r = reverseList(head?.next) else { return nil }
23         if r.next == nil {
24             r.next = head
25         } else {
26             head!.next!.next = head
27         }
28         head!.next = nil
29         return r
30     }
31 }

24ms

 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 reverseList(_ head: ListNode?) -> ListNode? {
14         if head == nil || head?.next == nil {
15             return head
16         }
17         
18         guard let r = reverseList(head?.next) else { return nil }
19         if r.next == nil {
20             r.next = head
21         } else {
22             head!.next!.next = head
23         }
24         head!.next = nil
25         return r
26     }
27 }

24ms

 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 reverseList(_ head: ListNode?) -> ListNode? {
14         if head == nil {
15             return nil
16         }
17         var root = head
18         
19         var stack = [ListNode]()
20         while root != nil {
21             stack.append(root!)
22             root = root!.next
23         }
24         
25         root = stack.last!
26         var next = root
27         for i in stride(from: stack.count - 2, to: -1, by: -1) {
28             let node = stack[i]
29             node.next = nil
30             next?.next = node
31             next = node;
32         }
33         return root
34     }
35 }
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