In this post, you will know how to solve the Reverse Linked List II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.Reverse Linked List II Leetcode SolutionsOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.ProblemGiven 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 <= 5001 <= left <= right <= nReverse Linked List II Leetcode Solutions in Pythonclass Solution: # @param head, a ListNode # @param m, an integer # @param n, an integer # @return a ListNode def reverseBetween(self, head, m, n): if m == n: return head dummyNode = ListNode(0) dummyNode.next = head pre = dummyNode for i in range(m - 1): pre = pre.next # reverse the [m, n] nodes reverse = None cur = pre.next for i in range(n - m + 1): next = cur.next cur.next = reverse reverse = cur cur = next pre.next.next = cur pre.next = reverse return dummyNode.nextReverse Linked List II Leetcode Solutions in CPPclass Solution { public: ListNode* reverseBetween(ListNode* head, int m, int n) { ListNode *dummy = new ListNode(0), *pre = dummy, *cur; dummy -> next = head; for (int i = 0; i < m - 1; i++) { pre = pre -> next; } cur = pre -> next; for (int i = 0; i < n - m; i++) { ListNode* temp = pre -> next; pre -> next = cur -> next; cur -> next = cur -> next -> next; pre -> next -> next = temp; } return dummy -> next; } };Reverse Linked List II Leetcode Solutions in Javapublic ListNode reverseBetween(ListNode head, int m, int n) { if(head == null) return null; ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list dummy.next = head; ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing for(int i = 0; i<m-1; i++) pre = pre.next; ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed ListNode then = start.next; // a pointer to a node that will be reversed // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3 // dummy-> 1 -> 2 -> 3 -> 4 -> 5 for(int i=0; i<n-m; i++) { start.next = then.next; then.next = pre.next; pre.next = then; then = start.next; } // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4 // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish) return dummy.next; }Note: This problem Reverse Linked List II is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: Decode Ways Leetcode Solution Post navigationRestore IP Addresses Leetcode Solution Decode Ways Leetcode Solution