# Sublist Search (Search a linked list in another list)

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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;

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;
}
}

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" :

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.

# GATE CS Corner    Company Wise Coding Practice

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