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Merge two sorted linked lists

Tags : linkedlist, geeksforgeeks, cpp, easy

Problem Statement - link #

Given two sorted linked lists consisting of N and M nodes respectively. The task is to merge both of the list (in-place) and return head of the merged list.

Your Task: The task is to complete the function sortedMerge() which takes references to the heads of two linked lists as the arguments and returns the head of merged linked list.

Expected Time Complexity: O(N+M)
Expected Auxiliary Space: O(1)

Examples #

Example 1:

Input:
N = 4, M = 3 
valueN[] = {5,10,15,40}
valueM[] = {2,3,20}
Output: 2 3 5 10 15 20 40
Explanation: After merging the two linked
lists, we have merged list as 2, 3, 5,
10, 15, 20, 40.

Example 2:

Input:
N = 2, M = 2
valueN[] = {1,1}
valueM[] = {2,4}
Output:1 1 2 4
Explanation: After merging the given two
linked list , we have 1, 1, 2, 4 as
output.

Constraints #

Solutions #

class Solution{
    public:
    /*
    struct Node {
    int data;
    struct Node *next;
    Node(int x) {
        data = x;
        next = NULL;
    }
    };
    */

    //Function to merge two sorted linked list.
    Node* sortedMerge(Node* head1, Node* head2)  
    {  
        // code here
        Node* head, *tail;
        if(head1->data <= head2->data) {
            head = tail = head1; head1 = head1->next;
        }else { head = tail = head2; head2 = head2->next; }
        while(head1!=NULL && head2!=NULL){
            if(head1->data <= head2->data){
                tail->next = head1;
                tail = head1;
                head1 = head1->next;
            }else {
                tail->next = head2;
                tail = head2;
                head2 = head2->next;
            }
        }
        if(head1==NULL) tail->next = head2;
        else tail->next = head1;
        return head;
    }  
};