Leetcode 328 Odd Even Linked List

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博客围绕单链表问题展开，要求将单链表中奇数结点和偶数结点分组，奇数在前，且要在原链表上操作，满足O(1)空间复杂度和O(nodes)时间复杂度。解题思路是构建一个偶数链表，用两个指针操作，最后将其链接到奇数链表末尾。

问题描述

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.

注意是交换的结点序的奇数偶数而不是结点的值的奇偶数

Example 1:

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

Example 2:（令第1,3,5...奇数结点在前）

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 ...

解题思路

由于第一个结点认为是odd的，只需要再构建一个even的链表，把even链表提取出来，最后再链接到odd的末尾即可。

构建even链表则需要两个指针，一个even头指针，一个指向even链表尾，可以保证将结点加到链表尾部。

题解

    public static ListNode oddEvenList(ListNode head) {
        if(head == null) return null;
        // 相当于odd链表的尾指针
        ListNode odd = head;
        // even 链表的尾指针
        ListNode even = head.next;
        // even 链表的头指针
        ListNode evenhead = even;

        while(even != null && even.next!=null){
            odd.next = odd.next.next;
            even.next = even.next.next;
            odd = odd.next;
            even = even.next;
        }
        odd.next = evenhead;
        return head;
    }