92. Reverse Linked List II LeetCode Solution

In this guide, you will get 92. Reverse Linked List II LeetCode Solution with the best time and space complexity. The solution to Reverse Linked List II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Reverse Linked List II solution in C++
4. Reverse Linked List II solution in Java
5. Reverse Linked List II solution in Python
6. Additional Resources

Problem Statement of Reverse Linked List II

Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.

Example 1:

Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]

Example 2:

Input: head = [5], left = 1, right = 1
Output: [5]

Constraints:

The number of nodes in the list is n.
1 <= n <= 500
-500 <= Node.val <= 500
1 <= left <= right <= n

Follow up: Could you do it in one pass?

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(n)

92. Reverse Linked List II LeetCode Solution in C++

class Solution {
 public:
  ListNode* reverseBetween(ListNode* head, int left, int right) {
    if (left == 1)
      return reverseN(head, right);

    head->next = reverseBetween(head->next, left - 1, right - 1);

    return head;
  }

 private:
  ListNode* reverseN(ListNode* head, int n) {
    if (n == 1)
      return head;

    ListNode* newHead = reverseN(head->next, n - 1);
    ListNode* headNext = head->next;
    head->next = headNext->next;
    headNext->next = head;

    return newHead;
  }
};
/* code provided by PROGIEZ */

92. Reverse Linked List II LeetCode Solution in Java

class Solution {
  public ListNode reverseBetween(ListNode head, int left, int right) {
    if (left == 1)
      return reverseN(head, right);

    head.next = reverseBetween(head.next, left - 1, right - 1);

    return head;
  }

  private ListNode reverseN(ListNode head, int n) {
    if (n == 1)
      return head;

    ListNode newHead = reverseN(head.next, n - 1);
    ListNode headNext = head.next;
    head.next = headNext.next;
    headNext.next = head;

    return newHead;
  }
}
// code provided by PROGIEZ

92. Reverse Linked List II LeetCode Solution in Python

class Solution:
  def reverseBetween(
      self,
      head: ListNode | None,
      left: int,
      right: int,
  ) -> ListNode | None:
    if left == 1:
      return self.reverseN(head, right)

    head.next = self.reverseBetween(head.next, left - 1, right - 1)
    return head

  def reverseN(self, head: ListNode | None, n: int) -> ListNode | None:
    if n == 1:
      return head

    newHead = self.reverseN(head.next, n - 1)
    headNext = head.next
    head.next = headNext.next
    headNext.next = head
    return newHead
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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