Reverse Nodes in k-Group leetcode java

題目：

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

 

題解：

這道題主要是利用reverse連結串列的方法，reverse的方法就是維護三個指標，然後別忘了儲存next指標就行。

程式碼如下：

 1     //http://www.cnblogs.com/lichen782/p/leetcode_Reverse_Nodes_in_kGroup.html
 2     /**
 3      * Reverse a link list between pre and next exclusively
 4      * an example:
 5      * a linked list:
 6      * 0->1->2->3->4->5->6
 7      * |           |   
 8      * pre        next
 9      * after call pre = reverse(pre, next)
10      * 
11      * 0->3->2->1->4->5->6
12      *          |  |
13      *          pre next
14      * @param pre 
15      * @param next
16      * @return the reversed list's last node, which is the precedence of parameter next
17      */
18     private static ListNode reverse(ListNode pre, ListNode next){
19         ListNode last = pre.next;//where first will be doomed "last"
20         ListNode cur = last.next;
21         while(cur != next){
22             last.next = cur.next;
23             cur.next = pre.next;
24             pre.next = cur;
25             cur = last.next;
26         }
27         return last;
28     }
29     
30     public static ListNode reverseKGroup(ListNode head, int k) {
31             if(head == null || k == 1)
32                 return head;
33                 
34             ListNode dummy = new ListNode(0);
35             dummy.next = head;
36             int count = 0;
37             ListNode pre = dummy;
38             ListNode cur = head;
39             while(cur != null){
40                 count ++;
41                 ListNode next = cur.next;
42                 if(count == k){
43                     pre = reverse(pre, next);
44                     count = 0;   
45                 }
46                 cur = next;
47             }
48          return dummy.next;
49         }

 Reference：http://codeganker.blogspot.com/2014/02/reverse-nodes-in-k-group-leetcode.html