328. Odd Even Linked List

题目：

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 ...

思路：

由于题目要求O(1)的空间复杂度，因此不能使用辅助的结构，只能在原来的链表上进行修改。

因此想到分别用两个指针指向奇数链头部和偶数链头部，最后再将偶数链连在奇数链的后面。

代码：

/**
* 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) {
//处理特殊情况，只有一个节点和只有两个节点的情况
if(head == null || head.next == null || head.next.next == null){
return head;
}
//用两个指针分别指向前两个节点，每次循环指向隔一个的值
ListNode odd = head;
ListNode even = head.next;
ListNode evenHead = head.next;
while(odd.next != null){
if(odd.next.next == null){break;}
ListNode t = odd.next.next;
odd.next = t;
odd = odd.next;

t = even.next.next;
even.next = t;
even = even.next;
}
//最后将偶数链表链在奇数链表之后
odd.next = evenHead;
return head;
}
}