DSA swap node pair leetcode problem error

Question

public ListNode swapPairs(ListNode node) {
    ListNode head = node;
    ListNode cur = node;
    while (head != null && head.next != null) {
        head = reverseList(head, 0);
        head.next = cur.next.next;
        head = head.next;
        cur = head;
    }
    return node;
}

public ListNode reverseList(ListNode node, int counter) {
    if (counter == 1 || node == null || node.next == null) {
        return node;
    } else {
        ListNode newHead = reverseList(node.next, counter++);
        node.next.next = node;
        node.next = null;
        return newHead;
    }
}

So this is a leetcode problem.

for example, if the given linked list is 1->2->3->4->5

then the output should be 2->1->4->3->5.

I saw the correct solution but i want to know why this code is not working. Can anyone please help me to understand this?

Do not tag this as dsa. Did you read what the dsa tag is for before you applied it to your question? — PaulMcKenzie
– PaulMcKenzie, Commented Sep 15, 2024 at 5:27
I don't use Java but one way to improve the question is to revise it to be a minimal reproducible example. Why do you need a recursive function to swap two nodes? Looking and input & output it seems you should swap node and next node, then advance 2 nodes. Rinse and repeat (i.e. in a loop). — user9706
– user9706, Commented Sep 15, 2024 at 6:23
Please explain your logic here: why would reversing the list help to achieve the goal? Also, how can return node be achieving the result, as it is still the first node in the list? For any list with two or more nodes that is already wrong. In those cases you should certainly not return the original node. Please clarify why you are surprised it doesn't work? — trincot
– trincot, Commented Sep 15, 2024 at 7:16

Arjit singh kushwaha · Accepted Answer · 2024-09-15 07:00:35Z

many errors are there , i wont be able to point all of them.

But let me tell you a concept.

let A be a linked list .

now when i do Node c = A.next

this mean c is point A next element right.

Now if i make c=null.

then A.next will also become null.

This is the concept of reference variables.

Meaning if you simply do c=a .then both c and a will point to same linked lint object.

now if you want to make a copy of a to c .you have to make new node and vopy elemnts of c to a .

Now.coming to the question posted here.

you can do two things basically.

either only change the value of linked listelements by swapping. or swap the nodes.

both should ideally work .

the problems with your code are- Here’s a simple breakdown of the errors in your code:

Wrong use of the reverseList function:

You're trying to swap pairs, but you're using a recursive list reversal function. This isn't needed since you just want to swap two nodes, not reverse an entire section.

counter++ issue in reverseList:

counter++ is post-increment, meaning it increases the value after it's used in the recursive call, which prevents the stopping condition from working properly.

Incorrect pointer handling after swapping:

After calling reverseList, you're not properly connecting the rest of the list. The line head.next = cur.next.next; can break the list structure, causing wrong behavior.

Unnecessary complexity:

You're using recursion and a separate function to reverse nodes, but you can directly swap adjacent nodes without extra complexity.

In short, the main errors are incorrect recursion, improper list handling, and using a reversal function when you only need to swap two nodes.

I ask that you correct the typos in your answer, for example vopy elemnts should be copy elements, correct? I also ask that you use markdown to better format your answer.

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