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

k = 2, 返回 2->1->4->3->5
k = 3, 返回 3->2->1->4->5

# 分析

head -> n1 -> n2 … nk -> nk+1 => head -> nk -> nk-1 .. n1 -> nk+1
return n1

# 代码

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode dummy = new ListNode(0);

while (true) {
break;
}
}

return dummy.next;
}

// head -> n1 -> n2 ... nk -> nk+1
// =>
// head -> nk -> nk-1 .. n1 -> nk+1
// return n1
public ListNode reverseK(ListNode head, int k) {
for (int i = 0; i < k; i++) {
if (nk == null) {
return null;
}
nk = nk.next;
}

if (nk == null) {
return null;
}

// reverse
ListNode nkplus = nk.next;

ListNode prev = null;
ListNode curt = n1;
while (curt != nkplus) {
ListNode temp = curt.next;
curt.next = prev;
prev = curt;
curt = temp;
}

// connect