LeetCode 206: Reverse Linked List

LeetCode 206: Reverse Linked List

    • [1. Problem Link 🔗](#1. Problem Link 🔗)
    • [2. Solution Overview 🧭](#2. Solution Overview 🧭)
    • [3. Solution 1: Iterative Approach (Recommended)](#3. Solution 1: Iterative Approach (Recommended))
      • [3.1. Algorithm](#3.1. Algorithm)
      • [3.2. Important Points](#3.2. Important Points)
      • [3.3. Java Implementation](#3.3. Java Implementation)
      • [3.4. Time & Space Complexity](#3.4. Time & Space Complexity)
    • [4. Solution 2: Recursive Approach](#4. Solution 2: Recursive Approach)
      • [4.1. Algorithm](#4.1. Algorithm)
      • [4.2. Important Points](#4.2. Important Points)
      • [4.3. Java Implementation](#4.3. Java Implementation)
      • [4.4. Time & Space Complexity](#4.4. Time & Space Complexity)
    • [5. Solution 3: Recursive with Helper Function](#5. Solution 3: Recursive with Helper Function)
      • [5.1. Algorithm](#5.1. Algorithm)
      • [5.2. Important Points](#5.2. Important Points)
      • [5.3. Java Implementation](#5.3. Java Implementation)
      • [5.4. Time & Space Complexity](#5.4. Time & Space Complexity)
    • [6. Solution 4: Iterative with Dummy Node](#6. Solution 4: Iterative with Dummy Node)
      • [6.1. Algorithm](#6.1. Algorithm)
      • [6.2. Important Points](#6.2. Important Points)
      • [6.3. Java Implementation](#6.3. Java Implementation)
      • [6.4. Time & Space Complexity](#6.4. Time & Space Complexity)
    • [7. Solution 5: Tail Recursive Approach](#7. Solution 5: Tail Recursive Approach)
      • [7.1. Algorithm](#7.1. Algorithm)
      • [7.2. Important Points](#7.2. Important Points)
      • [7.3. Java Implementation](#7.3. Java Implementation)
      • [7.4. Time & Space Complexity](#7.4. Time & Space Complexity)
    • [8. Solution Comparison 📊](#8. Solution Comparison 📊)
    • [9. Summary 📝](#9. Summary 📝)

LeetCode 206: Reverse Linked List

2. Solution Overview 🧭

Reverse a singly linked list. This is a fundamental linked list operation that appears frequently in interviews and real-world applications.

Example:

复制代码
Input: 1 → 2 → 3 → 4 → 5 → NULL
Output: 5 → 4 → 3 → 2 → 1 → NULL

Constraints:

  • The number of nodes in the list is the range [0, 5000]
  • -5000 <= Node.val <= 5000

Common approaches include:

  • Iterative Approach: Use three pointers to reverse links one by one
  • Recursive Approach: Recursively reverse the rest of the list and adjust pointers
  • Stack-based Approach: Use stack to reverse order (less efficient)

3. Solution 1: Iterative Approach (Recommended)

3.1. Algorithm

  • Use three pointers: prev, current, next
  • Traverse the list, reversing the next pointer of each node
  • Move pointers forward until the entire list is reversed

3.2. Important Points

  • Most efficient and intuitive
  • O(1) space complexity
  • Easy to understand and implement

3.3. Java Implementation

java 复制代码
/**
 * 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 reverseList(ListNode head) {
        ListNode prev = null;
        ListNode current = head;
        
        while (current != null) {
            ListNode nextTemp = current.next; // Store next node
            current.next = prev;              // Reverse the link
            prev = current;                   // Move prev forward
            current = nextTemp;               // Move current forward
        }
        
        return prev; // New head of reversed list
    }
}

3.4. Time & Space Complexity

  • Time Complexity: O(n)
  • Space Complexity: O(1)

4. Solution 2: Recursive Approach

4.1. Algorithm

  • Recursively reverse the rest of the list (head.next)
  • Adjust pointers so that the current node becomes the last node
  • The base case handles the end of the list

4.2. Important Points

  • Elegant but uses O(n) stack space
  • Good for understanding recursion
  • May cause stack overflow for very long lists

4.3. Java Implementation

java 复制代码
class Solution {
    public ListNode reverseList(ListNode head) {
        // Base case: empty list or single node
        if (head == null || head.next == null) {
            return head;
        }
        
        // Recursively reverse the rest of the list
        ListNode newHead = reverseList(head.next);
        
        // Adjust pointers: make current node the next of the last node
        head.next.next = head;
        head.next = null;
        
        return newHead;
    }
}

4.4. Time & Space Complexity

  • Time Complexity: O(n)
  • Space Complexity: O(n) - recursion stack

5. Solution 3: Recursive with Helper Function

5.1. Algorithm

  • Use a helper function to carry the previous node through recursion
  • Similar to iterative approach but implemented recursively
  • More intuitive for some developers

5.2. Important Points

  • Combines recursion with iterative logic
  • Easier to understand for some
  • Still uses O(n) stack space

5.3. Java Implementation

java 复制代码
class Solution {
    public ListNode reverseList(ListNode head) {
        return reverseHelper(head, null);
    }
    
    private ListNode reverseHelper(ListNode current, ListNode prev) {
        // Base case: reached end of list
        if (current == null) {
            return prev;
        }
        
        // Store next node before reversing
        ListNode nextTemp = current.next;
        
        // Reverse the current node
        current.next = prev;
        
        // Recursively process the next node
        return reverseHelper(nextTemp, current);
    }
}

5.4. Time & Space Complexity

  • Time Complexity: O(n)
  • Space Complexity: O(n) - recursion stack

6. Solution 4: Iterative with Dummy Node

6.1. Algorithm

  • Use a dummy node to simplify edge cases
  • Build the reversed list by inserting nodes at the beginning
  • More explicit but slightly more memory usage

6.2. Important Points

  • Handles edge cases gracefully
  • Clear separation of concerns
  • Good for learning purposes

6.3. Java Implementation

java 复制代码
class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null) return null;
        
        ListNode dummy = new ListNode(0);
        ListNode current = head;
        
        while (current != null) {
            ListNode nextTemp = current.next;
            
            // Insert current node at the beginning of reversed list
            current.next = dummy.next;
            dummy.next = current;
            
            current = nextTemp;
        }
        
        return dummy.next;
    }
}

6.4. Time & Space Complexity

  • Time Complexity: O(n)
  • Space Complexity: O(1)

7. Solution 5: Tail Recursive Approach

7.1. Algorithm

  • Optimized recursive approach that could benefit from tail call optimization
  • Some compilers can optimize this to use O(1) stack space
  • Clean functional programming style

7.2. Important Points

  • Potentially more efficient recursion
  • Clean mathematical approach
  • Java doesn't guarantee tail call optimization

7.3. Java Implementation

java 复制代码
class Solution {
    public ListNode reverseList(ListNode head) {
        return tailReverse(head, null);
    }
    
    private ListNode tailReverse(ListNode head, ListNode newHead) {
        if (head == null) {
            return newHead;
        }
        
        ListNode next = head.next;
        head.next = newHead;
        
        return tailReverse(next, head);
    }
}

7.4. Time & Space Complexity

  • Time Complexity: O(n)
  • Space Complexity: O(n) - though tail recursive, Java doesn't optimize

8. Solution Comparison 📊

Solution Time Complexity Space Complexity Advantages Disadvantages
Iterative O(n) O(1) Most efficient, intuitive None significant
Recursive O(n) O(n) Elegant, educational Stack overflow risk
Recursive Helper O(n) O(n) Combines recursion with iteration Still uses stack space
Iterative with Dummy O(n) O(1) Handles edges well Slight memory overhead
Tail Recursive O(n) O(n) Clean functional style No optimization in Java

9. Summary 📝

  • Key Insight: Reversing a linked list involves changing the direction of pointers while maintaining access to remaining nodes
  • Recommended Approach: Solution 1 (Iterative) for production code due to O(1) space
  • Recursive Insight: The recursive approach beautifully demonstrates divide-and-conquer by reversing the rest first
  • Pattern Recognition: This is a fundamental pattern for linked list manipulation

Recursive Thinking Process:

  1. Base case: empty list or single node is already reversed
  2. Recursively reverse everything after the current node
  3. Make the current node point to the new last node
  4. Make the last node point back to current node
  5. Break the original link to avoid cycles

For most practical purposes, the iterative approach is preferred, but understanding recursion is crucial for advanced algorithm problems.

相关推荐
Fluxart.ai18 分钟前
电商商品图审核怎么自动化?规则引擎、人工复核与发布门禁
java·前端·自动化
罗西的思考28 分钟前
【Agent OS / AIOS】AOHP 深度解读:当 OS 开始为 Agent 而设计
人工智能·算法·机器学习
民乐团扒谱机1 小时前
【微实验】组合优化matlab实战(马科维茨投资模型):在收益与风险之间,寻找最优的人生配比
大数据·人工智能·算法·机器学习·数学建模·matlab·组合优化
名字还没想好☜1 小时前
Java 线上内存泄漏排查实战:jmap 导堆、MAT 找 GC Roots 与四类常见泄漏
java·开发语言·jvm·内存泄漏
Nil2082 小时前
leetcode 160相交链表
算法·leetcode·链表
迷途之人不知返3 小时前
算法系列2:滑动窗口
算法
IT爱学堂3 小时前
尚硅谷 - 2025年3月Java+AI大模型应用开发
java·开发语言·人工智能
Herbert_hwt3 小时前
C语言零基础入门:循环控制与数据类型详解
c语言·数据结构·算法
小贤plus3 小时前
SpringBoot 三大核心注解精讲:@ControllerAdvice、@RestControllerAdvice、@Validated 分组校验(实战)
java
吠品4 小时前
Java byte数组与String互转:编码细节与踩坑记录
java·linux·服务器