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Data Structures & Algorithms June 14, 2020 2 min read

Leetcode 24 Swap Nodes in Pairs Java Solution

// SOLVING THIS WITH AN AI ASSISTANT (2026)

If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:

  1. Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
  2. Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
  3. Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
  4. Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”

The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.

Given a linked list, swap every two adjacent nodes and return its head.

You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Solution 1:

public class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null || head.next == null) return head;
        int count = 1;
           ListNode current = head;
           ListNode temp = head;
           ListNode returnNode = head.next;
           while(current.next!=null){
               if(getNextNode(current)!=null && count == 1){
               ListNode nextNode = getNextNode(current);
               current.next = nextNode.next;
               nextNode.next = current;
               temp = current;
               if(current.next!=null){
               current = current.next;
               }
               count++;
               }else if(getNextNode(current)!=null && count > 1){
                   ListNode nextNode = getNextNode(current);
               current.next = nextNode.next;
               nextNode.next = current;
               temp.next  = nextNode;
               temp = current;
               if(current.next!=null){
               current = current.next;
               }
               }
               
           }
           return returnNode;
               
    }
    
    public ListNode getNextNode(ListNode head){
        return head.next;
    }
}

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