7. Pairwise swap elements of a linked list

The problem can be found at the following link: Question Link

My Approach

I will traverse the linked list in pairs and swap the data values of adjacent nodes to achieve the pairwise swapping. The steps are as follows:

• Check if itr has a next node. If not, return the head as no swaps are possible.

• Initialize an iterator itr to the head of the list.

• Update the head to point to the second node (head = head->next) since it will become the new head after swapping.

• Update the next pointers of itr and the second node to perform the swap.

• Continue swapping pairs of nodes until the end of the list or until there's only one node left.

Time and Auxiliary Space Complexity

• Time Complexity: O(N) where N is the number of nodes in the linked list. We visit each node once.

• Auxiliary Space Complexity: O(1) as we use a constant amount of extra space.

Code (C++)

class Solution
{
    public:
    Node* pairWiseSwap(struct Node* head) 
    {
        if(!head->next)
            return head;

        struct Node* itr = head;
        head = head->next;
        itr->next = head->next;
        head->next = itr;
        while(itr){
            struct Node* curr = itr->next;
            if(curr && curr->next){
                itr->next = curr->next;
                curr->next = itr->next->next;
                itr->next->next = curr;
            }else
                break;
            itr = itr->next->next;
        }
        return head;
    }
};

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