LC_328. Odd Even Linked List

最新推荐文章于 2025-08-15 16:44:13 发布

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最新推荐文章于 2025-08-15 16:44:13 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/davidnyc/p/8457786.html

本文介绍了一种链表操作算法，该算法将链表中的节点按奇偶顺序重新排列，确保所有奇数位置节点位于偶数位置节点之前。算法采用原地修改的方式，不使用额外的空间，并且在O(n)时间内完成任务。文章通过具体示例展示了如何实现这一功能。

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https://leetcode.com/problems/odd-even-linked-list/description/
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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->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 ...

 1  public ListNode oddEvenList(ListNode head) {
 2         if (head == null || head.next == null) return head;
 3         ListNode oddCurr = head;
 4         ListNode evenHead = head.next ;
 5         ListNode evenCurr = head.next;
 6         /*
 7          1->2->3->4->5->NULL
 8          co ce
 9             eh
10          1->3     2->4
11               ->
12          * */
13         14         /*
15         even is always one node ahead so:
16          evenCurr != null == oddCurr.next != null
17          evenCurr.next != null == oddCurr.next.next != null
18 
19         重要！when evenCurr is at 4, evenCurr.next = 5 != null
20         */
21         while (evenCurr != null && evenCurr.next != null){
22             oddCurr.next = oddCurr.next.next ;
23             oddCurr = oddCurr.next;
24             //4->5.next=null 4->null
25             evenCurr.next = evenCurr.next.next ;
26             evenCurr = evenCurr.next;
27         }
28         //奇数结尾指向偶数的头
29         oddCurr.next = evenHead ;
30         return head;
31     }