328. Odd Even Linked List

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本文介绍了一种在O(1)空间复杂度和O(nodes)时间复杂度下，将单链表中的奇数节点和偶数节点进行分组的方法。通过这种算法，所有奇数节点会出现在所有偶数节点之前，同时保持原有的相对顺序。

328. 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 ...

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode oddEvenList(ListNode head) {
       if (head==null || head.next==null || head.next.next==null) return head;
         ListNode first0 = new ListNode(-1);
         ListNode second0 = new ListNode(-1);
         first0.next = head;//保留奇数的头结点
         second0.next = head.next;//保留偶数的头结点
         ListNode first = head;
         ListNode second = head.next;
         while (second!=null && second.next!=null) {
             first.next = second.next;
             first = first.next;
             second.next = first.next;
             second = second.next;
         }
         first.next = second0.next;//把偶数的头结点放到奇数的后面
         return first0.next;
     }
}