Reorder List

Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

分析：先用快慢指针找到链表的中点，然后翻转链表后半部分，再和前半部分组合。需要注意的是把链表分成两半时，前半段的尾节点要置为NULL，翻转链表时也要把尾节点置为NULL。

 1 /**

 2  * Definition for singly-linked list.

 3  * class ListNode {

 4  *     int val;

 5  *     ListNode next;

 6  *     ListNode(int x) {

 7  *         val = x;

 8  *         next = null;

 9  *     }

10  * }

11  */

12 public class Solution {

13     public void reorderList(ListNode head) {

14         // IMPORTANT: Please reset any member data you declared, as

15         // the same Solution instance will be reused for each test case.

16         if(head == null || head.next == null) return;

17         ListNode fast = head, slow = head;

18         while(fast != null && fast.next != null){

19             fast = fast.next.next;

20             slow = slow.next;

21         }

22         fast = slow.next;

23         slow.next = null;

24         fast = reverse(fast);

25         while(head != null && fast != null){//merge

26             ListNode tmp1 = head.next;

27             ListNode tmp2 = fast.next;

28             head.next = fast;

29             fast.next = tmp1;

30             head = tmp1;

31             fast = tmp2;

32         }

33     }

34     ListNode reverse(ListNode root){

35         ListNode first = null, second = root;

36         while(second != null){

37             ListNode tmp = second.next;

38             second.next = first;

39             first = second;

40             second = tmp;

41         }

42         return first;

43     }

44 }