Leetcode 206. Reverse Linked List

Similar Questions

Reverse Linked List II Binary Tree Upside Down Palindrome Linked List

思路：链表反转。

解法一：迭代。

添加头节点（推荐）：不断将当前元素start插入dummy和dummy.next之间，实现反转。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode reverseList(ListNode head) {
11         if(head == null) return head;
12         ListNode dummy = new ListNode(0);
13         dummy.next = head;
14         ListNode pre = head, start = head.next;
15         while(start != null) {
16             pre.next = start.next;
17             start.next = dummy.next;
18             dummy.next = start;
19             start = pre.next;
20         }
21         return dummy.next;
22     }
23 }

不添加头节点：

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode reverseList(ListNode head) {
11         ListNode p0, p1;
12         p0 = null;
13         p1 = head;
14         while(p1 != null) {//确保p1不为空
15             ListNode p2 = p1.next;
16             p1.next = p0;
17             p0 = p1;
18             p1 = p2;
19         }
20         return p0;
21     }
22 }

解法二：递归。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode reverseList(ListNode head) {
11         if(head == null || head.next == null) return head;//这个判断很重要
12         ListNode p1 = head.next;
13         ListNode p0 = reverseList(p1);
14         p1.next = head;
15         head.next = null;
16         return p0;
17     }
18 }

Next challenges: Reverse Linked List II Binary Tree Upside Down

转载于:https://www.cnblogs.com/Deribs4/p/8406523.html

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