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}.

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 }

 

posted on 2013-12-05 05:22  Step-BY-Step  阅读(239)  评论(0)  编辑  收藏  举报

导航