L25. Reverse Nodes in k-Group

LinkedList.

Problem

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.

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

Note:

• Only constant extra memory is allowed.

• You may not alter the values in the list's nodes, only nodes itself may be changed.

Solution:

• use for loop to find the range we need to reverse;

• next = cur.next, is a new part need to reverse;

  • therefore, head.next = reverseKGroup(next, k);

• newHead is the head node of the part that need to be reversed;

• use recursion to reverse the part

public ListNode reverseKGroup(ListNode head, int k){
   if(head == null) return head;
   ListNode cur = head;
   for(int i = 0; i < k-1; i++){
      cur = cur.next;
      if(cur == null) return head;
   }
   ListNode next = cur.next;
   ListNode newHead = reverse(head);
   head.next = reverseKGroup(next, k);
   return newHead;
}
public ListNode reverse(ListNode head){
   if(head == null || head.next == null) return head;
   ListNode reversedHead = reverse(head.next);
   head.next.next = head;
   head.next = null;
   return reversedHead;
}