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1 题目：
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
2 思路：
我自己想出来的。使用递归，每次只处理k个，递归处理剩下的。
3 代码：
public ListNode reverseKGroup(ListNode head, int k) { if(head == null){ return null; } ListNode fake=new ListNode(0); fake.next = head; ListNode l = head; int count = 0; while(l != null){ count++; l = l.next; } if(k>count) return head; ListNode after = null; l = head; ListNode pre = l.next; for(int i=0; i<k ; i++){ fake.next = l; l.next = after; after = l; l = pre; if(pre != null) pre = pre.next; } head.next=reverseKGroup(l,k); return fake.next; }
[leetcode 25]Reverse Nodes in k-Group
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原文地址：http://www.cnblogs.com/lingtingvfengsheng/p/4814630.html
