Reorder List leetcode java

題目：

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}.

 

題解：

題目要重新按照 L₀→L_n→L₁→L_n-1→L₂→L_n-2→…來排列，看例子1->2->3->4會變成1->4->2->3，拆開來看，是{1，2}和{4，3}的組合，而{4，3}是{3，4}的逆序。這樣問題的解法就出來了。

第一步，將連結串列分為兩部分。

第二步，將第二部分連結串列逆序。

第三步，將連結串列重新組合。

 

程式碼如下：

 

 1     public void reorderList(ListNode head) {
 2         if(head==null||head.next==null)
 3             return;
 4         
 5         ListNode slow=head, fast=head;
 6         ListNode firsthalf = head;
 7         while(fast.next!=null&&fast.next.next!=null){
 8             slow = slow.next;
 9             fast = fast.next.next;
10         }
11         
12         ListNode secondhalf = slow.next;
13         slow.next = null;
14         secondhalf = reverseOrder(secondhalf);
15  
16         while (secondhalf != null) {
17             ListNode temp1 = firsthalf.next;
18             ListNode temp2 = secondhalf.next;
19  
20             firsthalf.next = secondhalf;
21             secondhalf.next = temp1;        
22  
23             firsthalf = temp1;
24             secondhalf = temp2;
25         }
26         
27     }
28     
29     public static ListNode reverseOrder(ListNode head) {
30  
31         if (head == null || head.next == null)
32             return head;
33  
34         ListNode pre = head;
35         ListNode curr = head.next;
36  
37         while (curr != null) {
38             ListNode temp = curr.next;
39             curr.next = pre;
40             pre = curr;
41             curr = temp;
42         }
43  
44         // set head node's next
45         head.next = null;
46  
47         return pre;
48     }

 Reference：//http://www.programcreek.com/2013/12/in-place-reorder-a-singly-linked-list-in-java/