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
题意：奇数，偶数编号节点分开，
两个指针一个指向奇数开始，一个指向偶数开始，每次把first的下一个节点指向second的下一个节点。
一个关键点是 偶数个节点最后一个是偶数，奇数节点最后一个是奇数，那么用一个count进行奇数，如果最后是奇数，那么要用second指向head2
如果是偶数，first指向head2.
最后返回head1。（在纸上画一下就明白了）具体看代码
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution { 10 public ListNode oddEvenList(ListNode head) { 11 int count=2; 12 if((head==null)||(null==head.next))return head; 13 ListNode head1=head,head2=head.next,first=head,second=head.next; 14 while(second.next!=null){ 15 first.next=second.next; 16 first=second; 17 second=second.next; 18 count++; 19 } 20 if(count%2==0){ 21 first.next=head2; 22 return head1; 23 }else{ 24 first.next=null; 25 second.next=head2; 26 return head1; 27 } 28 } 29 }
原文：http://www.cnblogs.com/pkuYang/p/5135542.html