328. Odd Even Linked List

标签：ase   nbsp   public   time   input   relative   ups   and   pac   

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 ...

含义：将列表中奇数位置的节点靠前，偶数位置的节点靠后

 1     public ListNode oddEvenList(ListNode head) {
 2         if (head != null) {
 3             ListNode odd = head, even = head.next, evenHead = even;
 4             while (even != null && even.next != null) {
 5                 odd.next = odd.next.next;
 6                 even.next = even.next.next;
 7                 odd = odd.next;
 8                 even = even.next;
 9             }
10             odd.next = evenHead;
11         }
12         return head;        
13     }

328. Odd Even Linked List

标签：ase   nbsp   public   time   input   relative   ups   and   pac   

原文地址：http://www.cnblogs.com/wzj4858/p/7729417.html