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: First linked list: 1->2->3->4->6 Second linked list be 2->4->6->8, Output: 2->4->6. The elements 2, 4, 6 are common in both the list so they appear in the intersection list. Input: First linked list: 1->2->3->4->5 Second linked list be 2->3->4, 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 program to implement// the above approachclass 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 program to implement// the above approachimport 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<Integer> set = new HashSet<Integer>();
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!