25.reverse-nodes-in-k-groups-en

Problem

https://leetcode.com/problems/reverse-nodes-in-k-group/

Problem Description

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

Traverse linked list from left to right, during traverse, group nodes in k, then reverse each group. How to reverse a linked list given start, end node?

Reverse linked list:

Initial a prev node null
For each move, use temp node to keep current next node.
During traverse, update current node pointing to previous node, update previous pointing to current node
Update current to temp

ListNode temp = curr.next;
curr.next = prev;
prev = curr;
curr = temp;

For example(as below pic): reverse the whole linked list 1->2->3->4->null -> 4->3->2->1->null

Here Reverse each group(k nodes):

First group, use count keep track linked list counts when traverse linked list
Use start to keep track each group start node position.
Use end to keep track each group end node position
Reverse（k nodes）AKA: (start, end) - start and end exclusively.
After reverse, update start point to reversed group last node.
If counts % k != 0, then end move to next(end=end.next), for each movecount+1.

As below pic show steps 4 and 5, reverse linked list in range （start， end）:

For example(as below pic)，head=[1,2,3,4,5,6,7,8], k = 3

NOTE: Usually we create a dummy node to solve linked list problem, because head node may be changed during operation. for example: here head updated from 1->3, and dummy (List(0)) keep the same.

Complexity Analysis

Time Complexity: O(n) - n is number of Linked List
Space Complexity: O(1)

Key Points

create a dummy node, dummy = ListNode(0)
Group linked list as k=3, keep track of start and end node for each group.
Reverse each group, update start and end node references
return dummy.next.

Code (`Java/Python3`)

Java Code

class ReverseKGroupsLinkedList {
  public ListNode reverseKGroup(ListNode head, int k) {
      if (head == null || k == 1) {
        return head;
      }
      ListNode dummy = new ListNode(0);
      dummy.next = head;

      ListNode start = dummy;
      ListNode end = head;
      int count = 0;
      while (end != null) {
        count++;
        // group
        if (count % k == 0) {
          // reverse linked list (start, end]
          start = reverse(start, end.next);
          end = start.next;
        } else {
          end = end.next;
        }
      }
      return dummy.next;
    }

     /** 
     * reverse linked list from range (start, end), return last node.
     * for example: 
     * 0->1->2->3->4->5->6->7->8
     * |           |
     * start       end
     * 
     * After call start = reverse(start, end)
     * 
     * 0->3->2->1->4->5->6->7->8
     *          |  |
     *       start end
     *       
     * @return the reversed list's 'start' node, which is the precedence of node end
     */
    private ListNode reverse(ListNode start, ListNode end) {
      ListNode curr = start.next;
      ListNode prev = start;
      ListNode first = curr;
      while (curr != end){
        ListNode temp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = temp;
      }
      start.next = prev;
      first.next = curr;
      return first;
    }
}

Python3 Cose

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        if head is None or k < 2:
            return head
        dummy = ListNode(0)
        dummy.next = head
        start = dummy
        end = head
        count = 0
        while end:
            count += 1
            if count % k == 0:
                start = self.reverse(start, end.next)
                end = start.next
            else:
                end = end.next
        return dummy.next

    def reverse(self, start, end):
        prev, curr = start, start.next
        first = curr
        while curr != end:
            temp = curr.next
            curr.next = prev
            prev = curr
            curr = temp
        start.next = prev
        first.next = curr
        return first

References

Leetcode Discussion (yellowstone)

Extension

Require from right to left reverse nodes in k groups. (ByteDance Interview)
Example，1->2->3->4->5->6->7->8, k = 3,
From right to left, group as k=3:
- 6->7->8 reverse to 8->7->6，
- 3->4->5 reverse to 5->4->3.
- 1->2 only has 2 nodes, which less than k=3, do nothing.
  return： 1->2->5->4->3->8->7->6

Here, we pre-process linked list, reverse it first, then using Reverse nodes in K groups solution:

Reverse linked list
From left to right, reverse linked list group as k nodes.
Reverse step #2 linked list

For example:1->2->3->4->5->6->7->8, k = 3

Reverse linked list: 8->7->6->5->4->3->2->1
Reverse nodes in k groups: 6->7->8->3->4->5->2->1
Reverse step#2 linked list: 1->2->5->4->3->8->7->6