Reverse Linked List II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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In this post, you will find the solution for the Reverse Linked List II in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Reverse Linked List II– LeetCode Problem

Reverse Linked List II– LeetCode Problem

Problem:

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

Reverse Linked List II– LeetCode Solutions

Reverse Linked List II Solution in C++:

class Solution {
 public:
  ListNode* reverseBetween(ListNode* head, int m, int n) {
    if (!head || m == n)
      return head;

    ListNode dummy(0, head);
    ListNode* prev = &dummy;

    for (int i = 0; i < m - 1; ++i)
      prev = prev->next;  // point to the node before the sublist [m, n]

    ListNode* tail = prev->next;  // will be the tail of the sublist [m, n]

    // reverse the sublist [m, n] one by one
    for (int i = 0; i < n - m; ++i) {
      ListNode* cache = tail->next;
      tail->next = cache->next;
      cache->next = prev->next;
      prev->next = cache;
    }

    return dummy.next;
  }
};

Reverse Linked List II Solution in Java:

class Solution {
  public ListNode reverseBetween(ListNode head, int m, int n) {
    if (head == null || m == n)
      return head;

    ListNode dummy = new ListNode(0, head);
    ListNode prev = dummy;

    for (int i = 0; i < m - 1; ++i)
      prev = prev.next; // point to the node before the sublist [m, n]

    ListNode tail = prev.next; // will be the tail of the sublist [m, n]

    // reverse the sublist [m, n] one by one
    for (int i = 0; i < n - m; ++i) {
      ListNode cache = tail.next;
      tail.next = cache.next;
      cache.next = prev.next;
      prev.next = cache;
    }

    return dummy.next;
  }
}

Reverse Linked List II Solution in Python:

class Solution:
  def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
    if not head and m == n:
      return head

    dummy = ListNode(0, head)
    prev = dummy

    for _ in range(m - 1):
      prev = prev.next  # point to the node before the sublist [m, n]

    tail = prev.next  # will be the tail of the sublist [m, n]

    # reverse the sublist [m, n] one by one
    for _ in range(n - m):
      cache = tail.next
      tail.next = cache.next
      cache.next = prev.next
      prev.next = cache

    return dummy.next