Odd Even Linked List

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

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 …

题目本身不难理解，就是把奇数位置的放在前面，偶数位置的放在后面，注意，人家题目说的清清楚楚，不是数据域的奇偶。

直接看图吧：

按照图中所示，设置三个指针pPrev, p, pNext，交换这三个指针的位置，然后再做调整。

代码实现如下：

public class OddEvenLinkedList {    public static ListNode oddEvenList(ListNode head) {        if (head == null || head.next == null)            return head;        ListNode pPrev = head;        ListNode p = head.next;        ListNode pNext = p.next;        while (p != null && p.next != null) {            // Swap            p.next = pNext.next;            pNext.next = pPrev.next;            pPrev.next = pNext;            // move pointers            pPrev = pPrev.next;            p = p.next;            if (p != null)                pNext = p.next;            else                pNext = null;        }        return head;    }    public static void main(String[] args) {        ListNode head = new ListNode(1);        head.next = new ListNode(2);        head.next.next = new ListNode(3);        head.next.next.next = new ListNode(4);        head.next.next.next.next = new ListNode(5);        head.next.next.next.next.next = new ListNode(6);        head.next.next.next.next.next.next = new ListNode(7);        head.next.next.next.next.next.next.next = new ListNode(8);        head.next.next.next.next.next.next.next.next = new ListNode(9);        head.next.next.next.next.next.next.next.next.next = new ListNode(10);        RemoveDuplicatesFromSortedListII.printList(head);        System.out.println();        RemoveDuplicatesFromSortedListII.printList(oddEvenList(head));    }}

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