LeetCode 25. Reverse Nodes in k-Group

25. Reverse Nodes in k-Group

Difficulty:: Hard

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

Language: Java

/**
 * 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) {
            if (head == null || k <= 1) {
                  return head;
           }
            ListNode dummy = new ListNode(0);
            dummy.next = head;
            head = dummy;
            while (head != null) {
                  head = reverseNextKNodes(head, k);
           }
            return dummy.next;
     }
      // head -> n1 -> n2 -> ... -> nk -> nk+1
      // =>
      // head -> nk -> nk-1 -> ... -> n1 -> nk+1
      // return n1
      private ListNode reverseNextKNodes(ListNode head, int k) {
            int count = 0;
            ListNode nk = head;
            for (int i = 0; i < k; i++) {
                  nk = nk.next;
                  if (nk == null) {
                        return null;
                 }
           }
            ListNode n1 = head.next;
            ListNode prev = head;
            ListNode cur = head.next;
            while (prev != nk) {
                  ListNode tmp = cur.next;
                  cur.next = prev;
                  prev = cur;
                  cur = tmp;
           }
            head.next = prev;
            n1.next = cur;
            return n1;
     }
}