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.
Examples:
Input: list1 = 10->20
list2 = 5->10->20
Output : LIST FOUNDInput: list1 = 1->2->3->4
list2 = 1->2->1->2->3->4
Output: LIST FOUNDInput: list1 = 1->2->3->4
list2 = 1->2->2->1->2->3
Output: LIST NOT FOUND
Algorithm:
- Take first node of second list.
- Start matching the first list from this first node.
- If whole lists match return true.
- Else break and take first list to the first node again.
- And take second list to its second node.
- Repeat these steps until any of linked lists becomes empty.
- If first list becomes empty then list found else not.
Below is the implementation.
C++
// 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; } |
Java
// Java program to find a list in second list import java.util.*; class GFG { // A Linked List node static class Node { int data; Node next; }; // Returns true if first list is // present in second list static boolean 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 */static void printList(Node node) { while (node != null) { System.out.printf("%d ", node.data); node = node.next; } } // Function to add new node to linked lists static Node newNode(int key) { Node temp = new Node(); temp.data= key; temp.next = null; return temp; } // Driver Code public static void main(String[] args) { /* 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); if(findList(a, b) == true) System.out.println("LIST FOUND"); else System.out.println("LIST NOT FOUND"); } } // This code is contributed by Princi Singh |
Python3
# Python3 program to find a list in second list class Node: def __init__(self, value = 0): self.value = value self.next = None # Returns true if first list is # present in second list def findList(first, second): # If both linked lists are empty/None, # return True if not first and not second: return True # If ONLY one of them is empty, # return False if not first or not second: return False ptr1 = first ptr2 = second # Traverse the second LL by # picking nodes one by one while ptr2: # Initialize 'ptr2' with current # node of 'second' ptr2 = second # Start matching first LL # with second LL while ptr1: # If second LL become empty and # first not, return False, # since first LL has not been # traversed completely if not ptr2: return False # If value of both nodes from both # LLs are equal, increment pointers # for both LLs so that next value # can be matched else if ptr1.value == ptr2.value: ptr1 = ptr1.next ptr2 = ptr2.next # If a single mismatch is found # OR ptr1 is None/empty,break out # of the while loop and do some checks else: break # check 1 : # If 'ptr1' is None/empty,that means # the 'first LL' has been completely # traversed and matched so return True if not ptr1: return True # If check 1 fails, that means, some # items for 'first' LL are still yet # to be matched, so start again by # bringing back the 'ptr1' to point # to 1st node of 'first' LL ptr1 = first # And increment second node element to next second = second.next return False # Driver Code # Let us create two linked lists to # test the above functions. # Created lists would be be # node_a: 1->2->3->4 # node_b: 1->2->1->2->3->4 node_a = Node(1) node_a.next = Node(2) node_a.next.next = Node(3) node_a.next.next.next = Node(4) node_b = Node(1) node_b.next = Node(2) node_b.next.next = Node(1) node_b.next.next.next = Node(2) node_b.next.next.next.next = Node(3) node_b.next.next.next.next.next = Node(4) if findList(node_a, node_b): print("LIST FOUND") else: print("LIST NOT FOUND") # This code is contributed by GauriShankarBadola |
C#
// C# program to find a list in second list using System; class GFG { // A Linked List node class Node { public int data; public Node next; }; // Returns true if first list is // present in second list static Boolean 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 */static void printList(Node node) { while (node != null) { Console.Write("{0} ", node.data); node = node.next; } } // Function to add new node to linked lists static Node newNode(int key) { Node temp = new Node(); temp.data= key; temp.next = null; return temp; } // Driver Code public static void Main(String[] args) { /* 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); if(findList(a, b) == true) Console.Write("LIST FOUND"); else Console.Write("LIST NOT FOUND"); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program to find a // list in second list // A Linked List node class Node { constructor() { this.data = 0; this.next = null; } } // Returns true if first list is // present in second list function findList(first, second) { var 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 */ function printList(node) { while (node != null) { document.write("%d ", node.data); node = node.next; } } // Function to add new node to linked lists function newNode(key) { var temp = new Node(); temp.data = key; temp.next = null; return temp; } // Driver Code /* 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 */ var a = newNode(1); a.next = newNode(2); a.next.next = newNode(3); a.next.next.next = newNode(4); var 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); if (findList(a, b) == true) document.write("LIST FOUND"); else document.write("LIST NOT FOUND"); // This code contributed by gauravrajput1 </script> |
Output
LIST FOUND
Time Complexity: O(m*n) where m is the number of nodes in second list and n in first.
Auxiliary Space: O(1)
Optimization :
Above code can be optimized by using extra space i.e. stores the list into two strings and apply KMP algorithm. Refer https://ide.geeksforgeeks.org/3fXUaV for implementation provided by Nishant Singh
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