328.LeetCode_Linked List_Odd Even Linked List

最新推荐文章于 2024-11-05 16:27:01 发布

原创最新推荐文章于 2024-11-05 16:27:01 发布 · 456 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #java #linkedlist

leetcode 专栏收录该内容

31 篇文章

订阅专栏

本文介绍了一种在链表中按节点位置进行奇偶排序的方法，确保奇数位置的节点位于偶数位置节点之前，同时保持原有的相对顺序不变。算法采用O(1)的空间复杂度和O(n)的时间复杂度实现。

328. Odd Even Linked List

My Submissions

Total Accepted: 29029 Total Submissions: 76708 Difficulty: Medium

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 ...

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

https://leetcode.com/problems/odd-even-linked-list/

一、题目描述：

输入一个LinkedList，输出按奇偶位置排序，奇数位置在前，偶数位置在后。

二、JavaCode

public class Solution {
    public ListNode oddEvenList(ListNode head) {
       if(head == null || head.next == null)
            return head;
        ListNode odd = new ListNode(0);
        //偶数循环节点
        ListNode oddCurr = odd; 
        ListNode even = new ListNode(0);
        //基数循环节点
        ListNode evenCurr = even;
        int count = 0;
        while(head != null) 
        {
            count ++;   
            if(count % 2 == 0) 
            {
                evenCurr.next = new ListNode(head.val);
                evenCurr = evenCurr.next;
            } 
            else
            {
                oddCurr.next = new ListNode(head.val);
                oddCurr = oddCurr.next;
            }
            head = head.next;
        }
        
        oddCurr.next = even.next;
        return odd.next;
    }
}