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 list's nodes, only nodes themselves may be changed.
Example 1:
**Input:** head = [1,2,3,4,5], k = 2
**Output:** [2,1,4,3,5]
Example 2:
**Input:** head = [1,2,3,4,5], k = 3
**Output:** [3,2,1,4,5]
Example 3:
**Input:** head = [1,2,3,4,5], k = 1
**Output:** [1,2,3,4,5]
Example 4:
**Input:** head = [1], k = 1
**Output:** [1]
Constraints:
The number of nodes in the list is in the range sz.
1 <= sz <= 5000
0 <= Node.val <= 1000
1 <= k <= sz
Follow-up: Can you solve the problem in O(1) extra memory space?
ac1: non-recursive
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
// validate remain nodes > k
// cut k nodes
// reverse it
// dummy
// loop, remain, cut, ranger, step
// corner case
if (head == null || head.next == null || k <= 1) return head;
ListNode dum = new ListNode(0);
ListNode d = dum;
while (head != null) {
ListNode cut = head, ranger = head;
int step = 1;
// walk
while (ranger.next != null && step < k) {
ranger = ranger.next;
step++;
}
// validate
if (step < k) break;
// cut nodes
head = ranger.next;
ranger.next = null;
// add to dum
d.next = reverse(cut);
while (d.next != null) d = d.next;
}
// add the remain
d.next = head;
return dum.next;
}
private ListNode reverse(ListNode cut) {
ListNode prev = null;
while (cut != null) {
ListNode tmp = cut;
cut = cut.next;
tmp.next = prev;
prev = tmp;
}
return prev;
}
}
