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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 = 4Output:[1,4,3,2,5]

**Example 2:**

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