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
思路：
没啥思路，就是指针交换
package list; public class ReverseNodesInKGroup { public ListNode reverseKGroup(ListNode head, int k) { if (head == null || k < 2) return head; ListNode first = new ListNode(0); first.next = head; ListNode prevp = first; ListNode p = head; ListNode q = p; while (q != null) { int i = 0; for (; q != null && i < k - 1; ++i) { q = q.next; } if (i == k - 1 && q != null) { ListNode qNext = q.next; q.next = null; ListNode newHead = reverseList(p); prevp.next = newHead; prevp = p; p = qNext; q = qNext; prevp.next = qNext; } } return first.next; } // This function will only be invoked when the list contains more than one node. private ListNode reverseList(ListNode head) { ListNode prevp = null; ListNode p = head; ListNode q = head.next; while (p != null && q != null) { ListNode tmp = q.next; q.next = p; p.next = prevp; prevp = p; p = q; q = tmp; } return p; } public static void main(String[] args) { // TODO Auto-generated method stub ListNode a1 = new ListNode(1); ListNode a2 = new ListNode(2); ListNode a3 = new ListNode(3); ListNode a4 = new ListNode(4); ListNode a5 = new ListNode(5); a1.next = a2; a2.next = a3; a3.next = a4; a4.next = a5; a5.next = null; ReverseNodesInKGroup r = new ReverseNodesInKGroup(); ListNode x = r.reverseKGroup(a1, 3); while (x != null) { System.out.println(x.val); x = x.next; } } }