LeetCode-Odd Even Linked List_leetcode odd even linked list-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_24133491/article/details/96271090

本文介绍了一种链表操作算法，该算法可以将给定链表中的奇数位置节点移动到所有偶数位置节点之前，同时保持节点的原始顺序不变。通过定义三个关键指针并采用O(1)的空间复杂度和O(nodes)的时间复杂度来实现。

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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 1:

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

Example 2:

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

题意：给定一个链表，将位于奇数位的节点移动到所有偶数位节点之前，并且需要保持他们在原链表中的相对位置；

解法：首先，定义如下的三个变量

oddNext = head：指向排序后链表中奇数位节点的最后一个节点；
p = head.next：用于遍历链表节点；
pre = head：p的前一个节点

通过移动 $p$ 和 $p r e$ 节点，找到奇数位的节点，将其删除并移动到 $o d d N e x t$ 的尾部，同时更新 $o d d N e x t$ 节点；

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if (head == null || head.next == null || head.next.next == null) {
            return head;
        }
        
        ListNode pre = head;
        ListNode p = head.next;
        ListNode oddNext = head;
        while (p != null && p.next != null) {
            pre = pre.next;
            p = p.next;
            ListNode temp = p.next;
            pre.next = p.next;
            p.next = oddNext.next;
            oddNext.next = p;
            oddNext = p;
            p = temp;
        }
        
        return head;
    }
}