Easy-题目16：328. Odd Even Linked List-优快云博客

本文介绍了一种高效的单链表奇数节点与偶数节点重组算法，通过新开两个头结点实现奇偶节点分离与合并，确保了空间复杂度为O(1)，时间复杂度为O(nodes)。

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题目原文：
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.
题目大意：
给出一个单链表，把奇数节点放在偶数节点的前面。
题目分析：
新开两个头结点oddHead和evenHead，然后扫链表，交替的接在两个新头结点的后面，最后把odd链表的尾节点接到even链表的头结点上。
源码：（language：java）

public class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null)
            return head;
        ListNode oddHead = head,evenHead =head.next;
        ListNode prevOdd = oddHead,prevEven = evenHead;

        while(prevOdd.next != null && prevEven.next != null){
            prevOdd.next = prevEven.next;
            prevOdd = prevOdd.next;

            prevEven.next = prevOdd.next;
            prevEven = prevEven.next;
        }
        prevOdd.next = evenHead;

        return oddHead;
    }
}