Sublist Search (Search a linked list in another list)

1.5

Given two linked lists, the task is to check whether the first list is present in 2nd list or not.

Input  : list1 =  10->20
         list2  = 5->10->20
Output : LIST FOUND

Input  : list1 =  1->2->3->4
         list2  = 1->2->1->2->3->4
Output : LIST FOUND

Input  : list1 =  1->2->3->4
         list2  = 1->2->2->1->2->3
Output : LIST NOT FOUND

Source:http://qa.geeksforgeeks.org/9248/given-linked-list-have-check-whether-first-present-2nd-one-not

Algorithm:
1- Take first node of second list.
2- Start matching the first list from this first node.
3- If whole lists match return true.
4- Else break and take first list to the first node again.
5- And take second list to its second node.
6- Repeat these steps until any of linked lists becomes empty.
7- If first list becomes empty then list found else not.

Below is C++ implementation.

// C++ program to find a list in second list
#include <bits/stdc++.h>
using namespace std;

// A Linked List node
struct Node
{
    int data;
    Node* next;
};

// Returns true if first list is present in second
// list
bool findList(Node* first, Node* second)
{
    Node* ptr1 = first, *ptr2 = second;

    // If both linked lists are empty, return true
    if (first == NULL && second == NULL)
        return true;

    // Else If one is empty and other is not return
    // false
    if ( first == NULL ||
        (first != NULL && second == NULL))
        return false;

    // Traverse the second list by picking nodes
    // one by one
    while (second != NULL)
    {
        // Initialize ptr2 with current node of second
        ptr2 = second;

        // Start matching first list with second list
        while (ptr1 != NULL)
        {
            // If second list becomes empty and first
            // not then return false
            if (ptr2 == NULL)
                return false;

            // If data part is same, go to next
            // of both lists
            else if (ptr1->data == ptr2->data)
            {
                ptr1 = ptr1->next;
                ptr2 = ptr2->next;
            }

            // If not equal then  break the loop
            else break;
        }

        // Return true if first list gets traversed
        // completely that means it is matched.
        if (ptr1 == NULL)
            return true;

        // Initialize ptr1 with first again
        ptr1 = first;

        // And go to next node of second list
        second = second->next;
    }

    return false;
}

/* Function to print nodes in a given linked list */
void printList(Node* node)
{
    while (node != NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
}

// Function to add new node to linked lists
Node *newNode(int key)
{
    Node *temp = new Node;
    temp-> data= key;
    temp->next = NULL;
    return temp;
}

/* Driver program to test above functions*/
int main()
{
    /* Let us create two linked lists to test
    the above functions. Created lists shall be
        a: 1->2->3->4
        b: 1->2->1->2->3->4*/
    Node *a = newNode(1);
    a->next = newNode(2);
    a->next->next = newNode(3);
    a->next->next->next = newNode(4);

    Node *b = newNode(1);
    b->next = newNode(2);
    b->next->next = newNode(1);
    b->next->next->next = newNode(2);
    b->next->next->next->next = newNode(3);
    b->next->next->next->next->next = newNode(4);

    findList(a,b) ? cout << "LIST FOUND" :
                    cout << "LIST NOT FOUND";

    return 0;
}

Output:

LIST FOUND

Time Complexity : O(m*n) where m is the number of nodes in second list and n in first.

Optimization :
Above code can be optimized by using extra space i.e. stores the list into two strings and apply KMP algorithm. Refer http://code.geeksforgeeks.org/3fXUaV for implementation provided by Nishant Singh .

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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