# [LeetCode] 25. 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.

k is a positive integer and is less than or equal to the length of the linked 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

将链表分成k个一组，按顺序每次翻转一组。

对于每一组，先找到组内的第k个节点，如果找不到则说明长度小于k，无须翻转，返回null。

/**
* 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 < 2) {
}

ListNode dummy = new ListNode(0);

}

return dummy.next;
}

public ListNode reverseK(ListNode head, int k) {
for (int i = 0; i < k; i++) {
nk = nk.next;
if (nk == null) {
return null;
}
}

ListNode prev = null;
ListNode curr = n1;
while (prev != nk) {
ListNode temp = curr.next;
curr.next = prev;
prev = curr;
curr = temp;
}
n1.next = curr;
}