GeeksforGeeks App
Open App
Browser
Continue

# Java Program For Finding Intersection Of Two Sorted Linked Lists

Given two lists sorted in increasing order, create and return a new list representing the intersection of the two lists. The new list should be made with its own memory — the original lists should not be changed.

Example:

```Input:
Output: 2->4->6.
The elements 2, 4, 6 are common in
both the list so they appear in the
intersection list.

Input:
Output: 2->3->4
The elements 2, 3, 4 are common in
both the list so they appear in the
intersection list.```

Method 1: Using Dummy Node.
Approach:
The idea is to use a temporary dummy node at the start of the result list. The pointer tail always points to the last node in the result list, so new nodes can be added easily. The dummy node initially gives the tail a memory space to point to. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either ‘a’ or ‘b’ and adding it to the tail. When the given lists are traversed the result is in dummy. next, as the values are allocated from next node of the dummy. If both the elements are equal then remove both and insert the element to the tail. Else remove the smaller element among both the lists.

Below is the implementation of the above approach:

## Java

 `// Java program to implement``// the above approach``class` `GFG``{ ``    ``// Head nodes for pointing to``    ``// 1st and 2nd linked lists``    ``static` `Node a = ``null``, b = ``null``;``  ` `    ``// Dummy node for storing``    ``// intersection``    ``static` `Node dummy = ``null``;``  ` `    ``// Tail node for keeping track of``    ``// last node so that it makes easy``    ``// for insertion``    ``static` `Node tail = ``null``;``    ` `    ``// class - Node``    ``static` `class` `Node``    ``{``        ``int` `data;``        ``Node next;` `        ``Node(``int` `data)``        ``{``            ``this``.data = data;``            ``next = ``null``;``        ``}``    ``}``    ` `    ``// Function for printing the list``    ``void` `printList(Node start)``    ``{``        ``Node p = start;``        ``while` `(p != ``null``)``        ``{``            ``System.out.print(p.data + ``" "``);``            ``p = p.next;``        ``}``        ``System.out.println();``    ``}``    ` `    ``// Inserting elements into list``    ``void` `push(``int` `data)``    ``{``        ``Node temp = ``new` `Node(data);``        ``if``(dummy == ``null``)``        ``{``            ``dummy = temp;``            ``tail = temp;``        ``}``        ``else``        ``{``            ``tail.next = temp;``            ``tail = temp;``        ``}``    ``}``    ` `    ``// Function for finding intersection``    ``// and adding it to dummy list``    ``void` `sortedIntersect()``    ``{     ``        ``// Pointers for iterating``        ``Node p = a,q = b;``        ``while``(p != ``null`  `&&  q != ``null``)``        ``{``            ``if``(p.data == q.data)``            ``{``                ``// Add to dummy list``                ``push(p.data);``                ``p = p.next;``                ``q = q.next;``            ``}``            ``else` `if``(p.data < q.data)``                ``p = p.next;``            ``else``                ``q= q.next;``        ``}``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``GFG list = ``new` `GFG();``        ` `        ``// Creating first linked list``        ``list.a = ``new` `Node(``1``);``        ``list.a.next = ``new` `Node(``2``);``        ``list.a.next.next = ``new` `Node(``3``);``        ``list.a.next.next.next = ``new` `Node(``4``);``        ``list.a.next.next.next.next = ``new` `Node(``6``);` `        ``// Creating second linked list``        ``list.b = ``new` `Node(``2``);``        ``list.b.next = ``new` `Node(``4``);``        ``list.b.next.next = ``new` `Node(``6``);``        ``list.b.next.next.next = ``new` `Node(``8``);``        ` `        ``// Function call for intersection``        ``list.sortedIntersect();``      ` `        ``// Print required intersection``        ``System.out.println(``               ``"Linked list containing common items of a & b"``);``        ``list.printList(dummy);``    ``}``}``// This code is contributed by Likhita AVL`

Output:

```Linked list containing common items of a & b
2 4 6 ```

Complexity Analysis:

• Time Complexity: O(m+n) where m and n are number of nodes in first and second linked lists respectively.
Only one traversal of the lists are needed.
• Auxiliary Space: O(min(m, n)).
The output list can store at most min(m,n) nodes .

Method 2: Use Hashing

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `public` `class` `LinkedList``{``    ``Node head;``    ``static` `class` `Node``    ``{``        ``int` `data;``        ``Node next;           ``            ` `        ``Node(``int` `d) ``        ``{``            ``data = d; ``            ``next=``null``;                ``     ``}``}``public` `void` `printList()``{``    ``Node n = head;``    ``while``(n != ``null``)``    ``{``        ``System.out.println(n.data + ``" "``);``        ``n = n.next;``    ``}``}``     ` `public` `void` `append(``int` `d)``{``    ``Node n = ``new` `Node(d);``    ``if``(head== ``null``)``    ``{``        ``head = ``new` `Node(d);``        ``return``;``    ``}``            ` `    ``n.next = ``null``;``    ``Node last = head;``    ``while``(last.next !=``null``)``    ``{``        ``last = last.next;``    ``}``    ``last.next = n;``    ``return``;           ``}``        ` `static` `int``[] intersection(Node tmp1,``                          ``Node tmp2, ``int` `k)``{``    ``int``[] res = ``new` `int``[k];``    ``HashSet set = ``new` `HashSet();``    ``while``(tmp1 != ``null``)``    ``{               ``        ``set.add(tmp1.data);``        ``tmp1 = tmp1.next;``    ``}``            ` `    ``int` `cnt = ``0``;``            ` `    ``while``(tmp2 != ``null``)``    ``{``        ``if``(set.contains(tmp2.data))``        ``{``            ``res[cnt] = tmp2.data;``            ``cnt++;``        ``}``        ``tmp2 = tmp2.next;``    ``}``    ``return` `res;``            ` `}``     ` `// Driver code    ``public` `static` `void` `main(String[] args)``{``    ``LinkedList ll = ``new` `LinkedList();``    ``LinkedList ll1 = ``new` `LinkedList();``         ` `    ``ll.append(``0``);``    ``ll.append(``1``);``    ``ll.append(``2``);``    ``ll.append(``3``);``    ``ll.append(``4``);``    ``ll.append(``5``);``    ``ll.append(``6``);``    ``ll.append(``7``);        ``         ` `    ``ll1.append(``9``);``    ``ll1.append(``0``);``    ``ll1.append(``12``);``    ``ll1.append(``3``);``    ``ll1.append(``4``);``    ``ll1.append(``5``);``    ``ll1.append(``6``);``    ``ll1.append(``7``);``    ``int``[] arr= intersection(ll.head,``                            ``ll1.head,``6``);``    ``for``(``int` `i : arr)``    ``{``        ``System.out.println(i);``    ``}    ``}``// This code is contributed by ayyuce demirbas`

Output:

```0
3
4
5
6
7```

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(n) as using extra space

Please refer complete article on Intersection of two Sorted Linked Lists for more details!

My Personal Notes arrow_drop_up