Reverse Linked List II LeetCode Solution

Last updated on August 4th, 2024 at 11:43 pm

Here, We see Reverse Linked List II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Linked List

Level of Question

Medium

Reverse Linked List II LeetCode Solution

Reverse Linked List II LeetCode Solution

Problem Statement

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]

1. Reverse Linked List II Leetcode Solution C++

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
       ListNode *dummy = new ListNode(0), *pre = dummy, *cur;
       dummy -> next = head;
       for (int i = 0; i < left - 1; i++) {
           pre = pre -> next;
       }
       cur = pre -> next;
       for (int i = 0; i < right - left; i++) {
           ListNode* temp = pre -> next;
           pre -> next = cur -> next;
           cur -> next = cur -> next -> next;
           pre -> next -> next = temp;
       }
       return dummy -> next;        
    }
};

2. Reverse Linked List II Leetcode Solution Java

class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
    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<left-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

    for(int i=0; i<right-left; i++)
    {
        start.next = then.next;
        then.next = pre.next;
        pre.next = then;
        then = start.next;
    }
    return dummy.next;        
    }
}

3. Reverse Linked List II Leetcode Solution JavaScript

var reverseBetween = function(head, left, right) {
    let current = head, start = head, position = 1;
    
    while(position < left) {
        start = current
        current = current.next;
        position ++;
    }
    
    let reversedList = null,  tail = current;
    
    while(position >= left && position <= right) {
        const next = current.next;
        current.next = reversedList;
        reversedList = current;
        current = next;
        position ++
    }
    start.next = reversedList;
    tail.next = current;
    return left > 1 ? head : reversedList    
};

4. Reverse Linked List II Leetcode Solution Python

class Solution(object):
    def reverseBetween(self, head, left, right):
        if left == right:
            return head

        dummyNode = ListNode(0)
        dummyNode.next = head
        pre = dummyNode

        for i in range(left - 1):
            pre = pre.next
        
        reverse = None
        cur = pre.next
        for i in range(right - left + 1):
            next = cur.next
            cur.next = reverse
            reverse = cur
            cur = next

        pre.next.next = cur
        pre.next = reverse
        return dummyNode.next
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