【leetcode】Reorder List

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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}.

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题解：看起来最后变换得到的链表顺序很奇怪，起始很简单，就是把链表从中间折断，后半部分反转后和前半部分做归并即可。不过这个归并不是归并排序每次从两个链表中取较小的值，而是两个链表相互交错即可。

例如1->2->3->4，从中间折断得到两个链表1->2和3->4，后面一个链表反转得到4->3，然后和第一个链表交错归并得到1->3->2->4;

再例如1->2->3->4->5，从中间折断得到两个链表1->2->3和4->5，后一个链表反转得到5->4，和第一个链表交错归并得到1->4->2->5->3;

代码如下：

 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      private ListNode reverse(ListNode head){

14          ListNode newHead = null;

15          while(head != null){

16              ListNode temp = head.next;

17              head.next = newHead;

18              newHead = head;

19              head = temp;

20          }

21          return newHead;

22      }

23      

24      private void newMerge(ListNode head1,ListNode head2){

25          ListNode newHead = new ListNode(0);

26          

27          while(head1 != null && head2 != null){

28              newHead.next = head1;

29              head1 = head1.next;

30              newHead = newHead.next;

31              newHead.next = head2;

32              head2 = head2.next;

33              newHead = newHead.next;

34          }

35          if(head1 != null)

36              newHead.next = head1;

37          if(head2 != null)

38              newHead.next = head2;

39      }

40      private ListNode findMiddle(ListNode head){

41           ListNode slow = head;

42           ListNode fast = head.next;

43           while(fast != null && fast.next != null)

44           {

45               slow = slow.next;

46               fast = fast.next.next;

47           }

48           return slow;

49       }

50     public void reorderList(ListNode head) {

51         if(head == null || head.next == null)

52             return;

53         

54         ListNode mid = findMiddle(head);

55         ListNode tail = reverse(mid.next);

56         mid.next = null;

57         

58         newMerge(head, tail);

59     }

60 }