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C++ Program For Merging Two Sorted Linked Lists Such That Merged List Is In Reverse Order

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  • Last Updated : 22 Mar, 2022

Given two linked lists sorted in increasing order. Merge them such a way that the result list is in decreasing order (reverse order).

Examples: 

Input:  a: 5->10->15->40
        b: 2->3->20 
Output: res: 40->20->15->10->5->3->2

Input:  a: NULL
        b: 2->3->20 
Output: res: 20->3->2

A Simple Solution is to do following. 
1) Reverse first list ‘a’
2) Reverse second list ‘b’
3) Merge two reversed lists.
Another Simple Solution is first Merge both lists, then reverse the merged list.
Both of the above solutions require two traversals of linked list. 

How to solve without reverse, O(1) auxiliary space (in-place) and only one traversal of both lists? 
The idea is to follow merge style process. Initialize result list as empty. Traverse both lists from beginning to end. Compare current nodes of both lists and insert smaller of two at the beginning of the result list. 

1) Initialize result list as empty: res = NULL.
2) Let 'a' and 'b' be heads first and second lists respectively.
3) While (a != NULL and b != NULL)
    a) Find the smaller of two (Current 'a' and 'b')
    b) Insert the smaller value node at the front of the result.
    c) Move ahead in the list of the smaller nodes. 
4) If 'b' becomes NULL before 'a', insert all nodes of 'a' 
   into the result list at the beginning.
5) If 'a' becomes NULL before 'b', insert all nodes of 'a' 
   into result list at the beginning. 

Below is the implementation of above solution.

C++14




// C++ program to implement
// the above approach
#include<iostream>
using namespace std;
 
// Link list Node
struct Node
{
    int key;
    struct Node* next;
};
 
// Given two non-empty linked lists
// 'a' and 'b'
Node* SortedMerge(Node *a, Node *b)
{
    // If both lists are empty
    if (a==NULL && b==NULL)
        return NULL;
 
    // Initialize head of resultant
    // list
    Node *res = NULL;
 
    // Traverse both lists while both
    // of then have nodes.
    while (a != NULL && b != NULL)
    {
        // If a's current value is smaller
        // or equal to b's current value.
        if (a->key <= b->key)
        {
            // Store next of current Node
            // in first list
            Node *temp = a->next;
 
            // Add 'a' at the front of
            // resultant list
            a->next = res;
            res = a;
 
            // Move ahead in first list
            a = temp;
        }
 
        // If a's value is greater. Below steps
        // are similar to above (Only 'a' is
        // replaced with 'b')
        else
        {
            Node *temp = b->next;
            b->next = res;
            res = b;
            b = temp;
        }
    }
 
    // If second list reached end, but first
    // list has nodes. Add remaining nodes of
    // first list at the front of result list
    while (a != NULL)
    {
        Node *temp = a->next;
        a->next = res;
        res = a;
        a = temp;
    }
 
    // If first list reached end, but second
    // list has node. Add remaining nodes of
    // first list at the front of result list
    while (b != NULL)
    {
        Node *temp = b->next;
        b->next = res;
        res = b;
        b = temp;
    }
 
    return res;
}
 
/* Function to print Nodes in a
   given linked list */
void printList(struct Node *Node)
{
    while (Node!=NULL)
    {
        cout << Node->key << " ";
        Node = Node->next;
    }
}
 
/* Utility function to create a
   new node with given key */
Node *newNode(int key)
{
    Node *temp = new Node;
    temp->key = key;
    temp->next = NULL;
    return temp;
}
 
// Driver code
int main()
{
    // Start with the empty list
    struct Node* res = NULL;
 
    /* Let us create two sorted linked
       lists to test the above functions.
       Created lists shall be
       a: 5->10->15
       b: 2->3->20  */
    Node *a = newNode(5);
    a->next = newNode(10);
    a->next->next = newNode(15);
 
    Node *b = newNode(2);
    b->next = newNode(3);
    b->next->next = newNode(20);
 
    cout << "List A before merge: ";
    printList(a);
 
    cout << "List B before merge: ";
    printList(b);
 
    /* Merge 2 increasing order LLs
       in descresing order */
    res = SortedMerge(a, b);
 
    cout << "Merged Linked List is: ";
    printList(res);
 
    return 0;
}


Output: 

List A before merge: 
5 10 15 
List B before merge: 
2 3 20 
Merged Linked List is: 
20 15 10 5 3 2 

Time Complexity: O(N)

Auxiliary Space: O(1)

This solution traverses both lists only once, doesn’t require reverse and works in-place.
This article is contributed by Mohammed Raqeeb. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

Please refer complete article on Merge two sorted linked lists such that merged list is in reverse order for more details!


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