Problem Statement
Hard
Given the head of a linked list, reverse the nodes of the list k at a time, and return the 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]
Constraints:
- The number of nodes in the list is
n.
1 <= k <= n <= 50000 <= Node.val <= 1000
Follow-up: Can you solve the problem in O(1) extra memory space?
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode curr = head, kHead = curr, prev = null, dummy = new ListNode(0);
ListNode[] newChain = null, lastChain = new ListNode[] { dummy, dummy };
while (curr != null) {
kHead = curr;
int i = 0;
while (curr != null && i < k) {
prev = curr;
curr = curr.next;
i++;
}
if (i == k) {
prev.next = null; // disconnect next chain
newChain = reverseLL(kHead); // returns {head, tail} after reversal
lastChain[1].next = newChain[0]; // connect lastChain tail -> newChain head
lastChain = newChain;
} else
lastChain[1].next = kHead; // connect without reversal of remnants
}
return dummy.next;
}
/** Reverse a LinkedList
* @return ListNode[]{head, tail}
*/
private ListNode[] reverseLL(ListNode head) {
ListNode prev = null, curr = head;
while (curr != null) {
ListNode temp = curr.next;
curr.next = prev;
prev = curr;
curr = temp;
}
return new ListNode[] { prev, head }; // {newHead, newTail}
}
}