92. Reverse Linked List II

Medium

Problem Description

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.

Examples

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]
Jump to Solution: Python Java C++ JavaScript C#

Python Solution


# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
def reverseBetween(head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]:
    if not head or left == right:
        return head
    
    # Create dummy node to handle edge cases
    dummy = ListNode(0)
    dummy.next = head
    prev = dummy
    
    # Move to the position before reversal starts
    for _ in range(left - 1):
        prev = prev.next
    
    # Start reversal
    curr = prev.next
    for _ in range(right - left):
        next_node = curr.next
        curr.next = next_node.next
        next_node.next = prev.next
        prev.next = next_node
    
    return dummy.next

Java Solution


/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
        if (head == null || left == right) {
            return head;
        }
        
        // Create dummy node to handle edge cases
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        
        // Move to the position before reversal starts
        for (int i = 1; i < left; i++) {
            prev = prev.next;
        }
        
        // Start reversal
        ListNode curr = prev.next;
        for (int i = 0; i < right - left; i++) {
            ListNode nextNode = curr.next;
            curr.next = nextNode.next;
            nextNode.next = prev.next;
            prev.next = nextNode;
        }
        
        return dummy.next;
    }
}

C++ Solution


/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        if (!head || left == right) {
            return head;
        }
        
        // Create dummy node to handle edge cases
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* prev = dummy;
        
        // Move to the position before reversal starts
        for (int i = 1; i < left; i++) {
            prev = prev->next;
        }
        
        // Start reversal
        ListNode* curr = prev->next;
        for (int i = 0; i < right - left; i++) {
            ListNode* nextNode = curr->next;
            curr->next = nextNode->next;
            nextNode->next = prev->next;
            prev->next = nextNode;
        }
        
        ListNode* result = dummy->next;
        delete dummy;
        return result;
    }
};

JavaScript Solution


/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} left
 * @param {number} right
 * @return {ListNode}
 */
var reverseBetween = function(head, left, right) {
    if (!head || left === right) {
        return head;
    }
    
    // Create dummy node to handle edge cases
    const dummy = new ListNode(0);
    dummy.next = head;
    let prev = dummy;
    
    // Move to the position before reversal starts
    for (let i = 1; i < left; i++) {
        prev = prev.next;
    }
    
    // Start reversal
    let curr = prev.next;
    for (let i = 0; i < right - left; i++) {
        const nextNode = curr.next;
        curr.next = nextNode.next;
        nextNode.next = prev.next;
        prev.next = nextNode;
    }
    
    return dummy.next;
};

C# Solution


/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode ReverseBetween(ListNode head, int left, int right) {
        if (head == null || left == right) {
            return head;
        }
        
        // Create dummy node to handle edge cases
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = dummy;
        
        // Move to the position before reversal starts
        for (int i = 1; i < left; i++) {
            prev = prev.next;
        }
        
        // Start reversal
        ListNode curr = prev.next;
        for (int i = 0; i < right - left; i++) {
            ListNode nextNode = curr.next;
            curr.next = nextNode.next;
            nextNode.next = prev.next;
            prev.next = nextNode;
        }
        
        return dummy.next;
    }
}

Complexity Analysis

Solution Explanation

This solution uses iterative reversal with a dummy node:

Key points: