[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}.

 

基本思路：

目前想到的解法是，分三步来做：

1. 找出中间节点

2. 把中间节点之后的后半部分链表反序

3. 把前半部分链表及后半部分链表合并

 

Solution 1:

自己的代码，写得太乱了。。。很多地方可以改进的。。。

 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         if(head==null||head.next==null)
15             return;
16         ListNode fast=head;
17         ListNode slow=head;
18         while(fast.next!=null&&fast.next.next!=null){
19             fast=fast.next.next;
20             slow=slow.next;
21         }
22         ListNode head2=slow.next;
23         slow.next=null;
24         head2=reverseList(head2);
25         while(head!=null&&head2!=null){
26         ListNode temp=head.next;
27         head.next=head2;
28         ListNode temp2=head2.next;
29         head2.next=temp;
30         head=temp;
31         head2=temp2;  
32         }  
33     }
34 
35     private ListNode reverseList(ListNode head) {
36         
37         if(head==null||head.next==null)
38             return head;
39         // TODO Auto-generated method stub
40         ListNode dummy=new ListNode(-1);
41         dummy.next=head;
42         ListNode pre,l1,l2,l3;
43         pre=dummy;
44         l1=pre.next;
45         ListNode last=l1;
46         l2=l1.next;
47         l3=l2.next;
48         while(l3!=null){
49             l2.next=l1;
50             l1=l2;
51             l2=l3;
52             l3=l3.next;
53         }
54         dummy.next=l2;
55         l2.next=l1;
56         last.next=null;
57         return dummy.next;
58     }
59 }

 

Solution 2:

大神的代码：

 1 public class ReorderList {
 2     public void reorderList(ListNode head) {
 3         if (head == null || head.next == null)
 4             return;
 5         ListNode fast = head;
 6         ListNode late = head;
 7         while (fast.next != null && fast.next.next != null) {
 8             fast = fast.next.next;
 9             late = late.next;
10         }
11         ListNode ret = new ListNode(0);
12         ListNode cur = ret;
13         ListNode leftHalf = head;
14         ListNode rightHalf;
15         if (fast.next != null) {
16             rightHalf = reverseList(late.next);
17             late.next = null;
18         } else {
19             rightHalf = reverseList(late);
20             ListNode tmp = head;
21             while (tmp.next != late) {
22                 tmp = tmp.next;
23             }
24             tmp.next = null;
25         }
26         leftHalf = head;
27         while (leftHalf != null && rightHalf != null) {
28             cur.next = leftHalf;
29             leftHalf = leftHalf.next;
30             cur = cur.next;
31             cur.next = rightHalf;
32             rightHalf = rightHalf.next;
33             cur = cur.next;
34         }
35         if (leftHalf != null) {
36             cur.next = leftHalf;
37         } else if (rightHalf != null) {
38             cur.next = rightHalf;
39         }
40         head = ret.next;
41     }
42 
43     private ListNode reverseList(ListNode head) {
44         ListNode cur = head;
45         ListNode prev = null;
46         ListNode next = null;
47         while (cur != null) {
48             next = cur.next;
49             cur.next = prev;
50             prev = cur;
51             cur = next;
52         }
53         return prev;
54     }
55 }