LeetCode-328. Odd Even Linked List-优快云博客

本文介绍了一种在O(1)空间复杂度和O(nodes)时间复杂度下，将链表中的奇数节点和偶数节点按顺序重新排列的方法。通过分离奇数和偶数节点并最终合并的方式实现了这一目标。

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 ...

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

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问题是给出一个奇偶数相间的链表，返回一个前边是奇数后边是偶数的链表顺序和输入顺序一致，要求空间复杂度O(1)，时间复杂度O(nodes)。

思路：把奇数连接起来，偶数连接起来，最后连接奇偶。

实现代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    //1->2->3->4->5->NULL,
    public ListNode oddEvenList(ListNode head) {
        if(head == null)
            return null;
        ListNode odd = head;
        ListNode even = head.next;
        ListNode evenHead = head.next;
        while(odd.next != null && even.next !=null) {
            odd.next = even.next;
            even.next = even.next.next;
            odd = odd.next;
            even = even.next;
        }
        odd.next = evenHead;
        
        return head;
        
    }
}