Reverse Nodes in k-Group

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

题意：反转一个链表中的每k个节点，如果链表长度不能被k整除，则最后余数节点保持不变
解题思路：首先先对链表进行分组。然后在分组进行反转

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null || k <= 1) {
return head;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
//记录分组区间的开始节点
ListNode preNode = dummy, cross = head;
int count = 0;
while (cross != null) {
count++;
if (count % k == 0) {
//分组区间结束节点
preNode = reverse(preNode,cross.next);
cross = preNode.next;
} else {
cross = cross.next;
}
}
return dummy.next;
}
ListNode reverse(ListNode pre ,ListNode next){
ListNode last = pre.next;
ListNode curNode = last.next;
while (curNode != next) {
last.next = curNode.next;
curNode.next = pre.next;
pre.next = curNode;
curNode = last.next;
}
return last;
}

}