Intersection of two Sorted Linked Lists
Last Updated :
23 Feb, 2023
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:
C++
#include<bits/stdc++.h>
using namespace std;
struct Node {
int data;
Node* next;
};
void push(Node** head_ref, int new_data);
Node* sortedIntersect(Node* a, Node* b)
{
Node dummy;
Node* tail = &dummy;
dummy.next = NULL;
while (a != NULL && b != NULL) {
if (a->data == b->data) {
push((&tail->next), a->data);
tail = tail->next;
a = a->next;
b = b->next;
}
else if (a->data < b->data)
a = a->next;
else
b = b->next;
}
return (dummy.next);
}
void push(Node** head_ref, int new_data)
{
Node* new_node = (Node*)malloc(
sizeof(Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(Node* node)
{
while (node != NULL) {
cout << node->data <<" ";
node = node->next;
}
}
int main()
{
Node* a = NULL;
Node* b = NULL;
Node* intersect = NULL;
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
intersect = sortedIntersect(a, b);
cout<<"Linked list containing common items of a & b \n";
printList(intersect);
}
|
C
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
void push(struct Node** head_ref, int new_data);
struct Node* sortedIntersect(
struct Node* a,
struct Node* b)
{
struct Node dummy;
struct Node* tail = &dummy;
dummy.next = NULL;
while (a != NULL && b != NULL) {
if (a->data == b->data) {
push((&tail->next), a->data);
tail = tail->next;
a = a->next;
b = b->next;
}
else if (a->data < b->data)
a = a->next;
else
b = b->next;
}
return (dummy.next);
}
void push(struct Node** head_ref, int new_data)
{
struct Node* new_node = (struct Node*)malloc(
sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
int main()
{
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
intersect = sortedIntersect(a, b);
printf("\n Linked list containing common items of a & b \n ");
printList(intersect);
getchar();
}
|
Java
import java.util.*;
import java.io.*;
public class GFG
{
static Node a = null, b = null;
static Node dummy = null;
static Node tail = null;
static class Node {
int data;
Node next;
Node(int data) {
this.data = data;
next = null;
}
}
void printList(Node start) {
Node p = start;
while (p != null) {
System.out.print(p.data + " ");
p = p.next;
}
System.out.println();
}
void push(int data) {
Node temp = new Node(data);
if(dummy == null) {
dummy = temp;
tail = temp;
}
else {
tail.next = temp;
tail = temp;
}
}
void sortedIntersect()
{
Node p = a,q = b;
while(p != null && q != null)
{
if(p.data == q.data)
{
push(p.data);
p = p.next;
q = q.next;
}
else if(p.data < q.data)
p = p.next;
else
q= q.next;
}
}
public static void main(String args[])
{
GFG list = new GFG();
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);
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);
list.sortedIntersect();
System.out.println("Linked list containing common items of a & b");
list.printList(dummy);
}
}
|
Python3
class Node:
def __init__(self):
self.data = 0
self.next = None
def sortedIntersect(a, b):
dummy = Node()
tail = dummy;
dummy.next = None;
while (a != None and b != None):
if (a.data == b.data):
tail.next = push((tail.next), a.data);
tail = tail.next;
a = a.next;
b = b.next;
elif(a.data < b.data):
a = a.next;
else:
b = b.next;
return (dummy.next);
def push(head_ref, new_data):
new_node = Node()
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref
def printList(node):
while (node != None):
print(node.data, end=' ')
node = node.next;
if __name__=='__main__':
a = None;
b = None;
intersect = None;
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
intersect = sortedIntersect(a, b);
print("Linked list containing common items of a & b ");
printList(intersect);
|
C#
using System;
public class GFG
{
static Node dummy = null;
static Node tail = null;
public class Node
{
public int data;
public Node next;
public Node(int data)
{
this.data = data;
next = null;
}
}
Node a = null, b = null;
void printList(Node start)
{
Node p = start;
while (p != null)
{
Console.Write(p.data + " ");
p = p.next;
}
Console.WriteLine();
}
void push(int data)
{
Node temp = new Node(data);
if(dummy == null)
{
dummy = temp;
tail = temp;
}
else
{
tail.next = temp;
tail = temp;
}
}
void sortedIntersect()
{
Node p = a,q = b;
while(p != null && q != null)
{
if(p.data == q.data)
{
push(p.data);
p = p.next;
q = q.next;
}
else if(p.data < q.data)
p = p.next;
else
q= q.next;
}
}
public static void Main(String []args)
{
GFG list = new GFG();
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);
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);
list.sortedIntersect();
Console.WriteLine("Linked list containing common items of a & b");
list.printList(dummy);
}
}
|
Javascript
<script>
var a = null, b = null;
var dummy = null;
var tail = null;
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
function printList(start) {
var p = start;
while (p != null) {
document.write(p.data + " ");
p = p.next;
}
document.write();
}
function push(data) {
var temp = new Node(data);
if (dummy == null) {
dummy = temp;
tail = temp;
} else {
tail.next = temp;
tail = temp;
}
}
function sortedIntersect() {
var p = a, q = b;
while (p != null && q != null) {
if (p.data == q.data) {
push(p.data);
p = p.next;
q = q.next;
} else if (p.data < q.data)
p = p.next;
else
q = q.next;
}
}
a = new Node(1);
a.next = new Node(2);
a.next.next = new Node(3);
a.next.next.next = new Node(4);
a.next.next.next.next = new Node(6);
b = new Node(2);
b.next = new Node(4);
b.next.next = new Node(6);
b.next.next.next = new Node(8);
sortedIntersect();
document.write(
"Linked list containing common items of a & b<br/>"
);
printList(dummy);
</script>
|
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: Using Local References.
Approach: This solution is structurally very similar to the above, but it avoids using a dummy node Instead, it maintains a struct node** pointer, lastPtrRef, that always points to the last pointer of the result list. This solves the same case that the dummy node did — dealing with the result list when it is empty. If the list is built at its tail, either the dummy node or the struct node** “reference” strategy can be used.
Below is the implementation of the above approach:
C++14
#include <bits/stdc++.h>
struct Node {
int data;
struct Node* next;
};
void push(struct Node** head_ref,
int new_data);
struct Node* sortedIntersect(
struct Node* a,
struct Node* b)
{
struct Node* result = NULL;
struct Node** lastPtrRef = &result;
while (a != NULL && b != NULL) {
if (a->data == b->data) {
push(lastPtrRef, a->data);
lastPtrRef = &((*lastPtrRef)->next);
a = a->next;
b = b->next;
}
else if (a->data < b->data)
a = a->next;
else
b = b->next;
}
return (result);
}
void push(struct Node** head_ref,
int new_data)
{
struct Node* new_node = (struct Node*)malloc(
sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
int main()
{
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
intersect = sortedIntersect(a, b);
printf("\n Linked list containing common items of a & b \n ");
printList(intersect);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
void push(struct Node** head_ref,
int new_data);
struct Node* sortedIntersect(
struct Node* a,
struct Node* b)
{
struct Node* result = NULL;
struct Node** lastPtrRef = &result;
while (a != NULL && b != NULL) {
if (a->data == b->data) {
push(lastPtrRef, a->data);
lastPtrRef = &((*lastPtrRef)->next);
a = a->next;
b = b->next;
}
else if (a->data < b->data)
a = a->next;
else
b = b->next;
}
return (result);
}
void push(struct Node** head_ref,
int new_data)
{
struct Node* new_node = (struct Node*)malloc(
sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
int main()
{
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
intersect = sortedIntersect(a, b);
printf("\n Linked list containing common items of a & b \n ");
printList(intersect);
getchar();
}
|
Java
import java.util.*;
import java.io.*;
public class GFG {
static class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
};
static Node sortedIntersect(Node a, Node b)
{
Node result = new Node(0);
Node curr = result;
while (a != null && b != null) {
if (a.data == b.data) {
curr.next = new Node(a.data);
curr = curr.next;
a = a.next;
b = b.next;
}
else if (a.data < b.data)
a = a.next;
else
b = b.next;
}
result = result.next;
return result;
}
static Node push(Node head_ref, int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
static void printList(Node node)
{
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
}
public static void main(String[] args)
{
Node a = null;
Node b = null;
Node intersect = null;
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
intersect = sortedIntersect(a, b);
System.out.print("\n Linked list containing "
+ "common items of a & b \n ");
printList(intersect);
}
}
|
Python3
class Node:
def __init__(self, d):
self.data = d
self.next = None
def sortedIntersect(a, b):
result = Node(0)
curr = result
while (a != None and b != None):
if (a.data == b.data):
curr.next = Node(a.data)
curr = curr.next
a = a.next
b = b.next
elif (a.data < b.data):
a = a.next
else:
b = b.next
result = result.next
return result
def push(head_ref, new_data):
new_node = Node(new_data)
new_node.next = head_ref
head_ref = new_node
return head_ref
def printList(node):
while (node != None):
print(node.data, end=" ")
node = node.next
a = None
b = None
intersect = None
a = push(a, 6)
a = push(a, 5)
a = push(a, 4)
a = push(a, 3)
a = push(a, 2)
a = push(a, 1)
b = push(b, 8)
b = push(b, 6)
b = push(b, 4)
b = push(b, 2)
intersect = sortedIntersect(a, b)
print("Linked list containing " + "common items of a & b")
printList(intersect)
|
C#
using System;
public class GFG {
public class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
};
static Node sortedIntersect(Node a, Node b)
{
Node result = new Node(0);
Node curr = result;
while (a != null && b != null) {
if (a.data == b.data) {
curr.next = new Node(a.data);
curr = curr.next;
a = a.next;
b = b.next;
}
else if (a.data < b.data)
a = a.next;
else
b = b.next;
}
result = result.next;
return result;
}
static Node push(Node head_ref, int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
static void printList(Node node)
{
while (node != null) {
Console.Write(node.data + " ");
node = node.next;
}
}
public static void Main(String[] args)
{
Node a = null;
Node b = null;
Node intersect = null;
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
intersect = sortedIntersect(a, b);
Console.Write("\n Linked list containing "
+ "common items of a & b \n ");
printList(intersect);
}
}
|
Javascript
<script>
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
function printList(node) {
while (node != null) {
document.write(node.data + " ");
node = node.next;
}
}
function push(head_ref, new_data) {
var new_node = new Node(new_data);
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
function sortedIntersect(a, b){
var result = new Node(0);
var lastptrRef = result;
while(a != null && b != null){
if(a.data == b.data){
lastptrRef.next = new Node(a.data);
lastptrRef = lastptrRef.next;
a = a.next;
b = b.next;
}
else if(a.data < b.data)
a = a.next;
else
b = b.next;
}
result = result.next;
return result;
}
var a = null;
var b = null;
var intersect = null;
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
intersect = sortedIntersect(a, b);
document.write("Linked list containing common items of a & b");
printList(intersect);
</script>
|
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(max(m, n)).
The output list can store at most m+n nodes.
Method 3: Recursive Solution.
Approach:
The recursive approach is very similar to the above two approaches. Build a recursive function that takes two nodes and returns a linked list node. Compare the first element of both the lists.
- If they are similar then call the recursive function with the next node of both the lists. Create a node with the data of the current node and put the returned node from the recursive function to the next pointer of the node created. Return the node created.
- If the values are not equal then remove the smaller node of both the lists and call the recursive function.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
struct Node
{
int data;
struct Node* next;
};
struct Node* sortedIntersect(struct Node* a,
struct Node* b)
{
if (a == NULL || b == NULL)
return NULL;
if (a->data < b->data)
return sortedIntersect(a->next, b);
if (a->data > b->data)
return sortedIntersect(a, b->next);
struct Node* temp = (struct Node*)malloc(
sizeof(struct Node));
temp->data = a->data;
temp->next = sortedIntersect(a->next,
b->next);
return temp;
}
void push(struct Node** head_ref, int new_data)
{
struct Node* new_node = (struct Node*)malloc(
sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(struct Node* node)
{
while (node != NULL)
{
cout << " " << node->data;
node = node->next;
}
}
int main()
{
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
intersect = sortedIntersect(a, b);
cout << "\n Linked list containing "
<< "common items of a & b \n ";
printList(intersect);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
struct Node* sortedIntersect(
struct Node* a,
struct Node* b)
{
if (a == NULL || b == NULL)
return NULL;
if (a->data < b->data)
return sortedIntersect(a->next, b);
if (a->data > b->data)
return sortedIntersect(a, b->next);
struct Node* temp
= (struct Node*)malloc(
sizeof(struct Node));
temp->data = a->data;
temp->next = sortedIntersect(a->next, b->next);
return temp;
}
void push(struct Node** head_ref, int new_data)
{
struct Node* new_node
= (struct Node*)malloc(
sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
int main()
{
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
intersect = sortedIntersect(a, b);
printf("\n Linked list containing common items of a & b \n ");
printList(intersect);
return 0;
}
|
Java
import java.util.*;
import java.io.*;
public class GFG{
static class Node
{
int data;
Node next;
};
static Node sortedIntersect(Node a,
Node b)
{
if (a == null || b == null)
return null;
if (a.data < b.data)
return sortedIntersect(a.next, b);
if (a.data > b.data)
return sortedIntersect(a, b.next);
Node temp = new Node();
temp.data = a.data;
temp.next = sortedIntersect(a.next,
b.next);
return temp;
}
static Node push(Node head_ref, int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
static void printList(Node node)
{
while (node != null)
{
System.out.print(" " + node.data);
node = node.next;
}
}
public static void main(String[] args)
{
Node a = null;
Node b = null;
Node intersect = null;
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
intersect = sortedIntersect(a, b);
System.out.print("\n Linked list containing "
+ "common items of a & b \n ");
printList(intersect);
}
}
|
Python3
class Node:
def __init__(self):
self.data = 0
self.next = None
def sortedIntersect(a, b):
if (a == None or b == None):
return None
if (a.data < b.data):
return sortedIntersect(a.next, b);
if (a.data > b.data):
return sortedIntersect(a, b.next);
temp = Node();
temp.data = a.data;
temp.next = sortedIntersect(a.next, b.next);
return temp;
def push(head_ref, new_data):
new_node = Node()
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
def printList(node):
while (node != None):
print(node.data, end=" ")
node = node.next;
a = None
b = None
intersect = None
a = push(a, 6)
a = push(a, 5)
a = push(a, 4)
a = push(a, 3)
a = push(a, 2)
a = push(a, 1)
b = push(b, 8)
b = push(b, 6)
b = push(b, 4)
b = push(b, 2)
intersect = sortedIntersect(a, b)
print("\n Linked list containing " + "common items of a & b");
printList(intersect)
|
C#
using System;
public class GFG {
public class Node {
public int data;
public Node next;
};
static Node sortedIntersect(Node a, Node b) {
if (a == null || b == null)
return null;
if (a.data < b.data)
return sortedIntersect(a.next, b);
if (a.data > b.data)
return sortedIntersect(a, b.next);
Node temp = new Node();
temp.data = a.data;
temp.next = sortedIntersect(a.next, b.next);
return temp;
}
static Node push(Node head_ref, int new_data) {
Node new_node = new Node();
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
static void printList(Node node) {
while (node != null) {
Console.Write(" " + node.data);
node = node.next;
}
}
public static void Main(String[] args) {
Node a = null;
Node b = null;
Node intersect = null;
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
intersect = sortedIntersect(a, b);
Console.Write("\n Linked list containing " + "common items of a & b \n ");
printList(intersect);
}
}
|
Javascript
<script>
class Node {
constructor(){
this.data = 0;
this.next = null;
}
}
function sortedIntersect(a, b) {
if (a == null || b == null)
return null;
if (a.data < b.data)
return sortedIntersect(a.next, b);
if (a.data > b.data)
return sortedIntersect(a, b.next);
var temp = new Node();
temp.data = a.data;
temp.next = sortedIntersect(a.next, b.next);
return temp;
}
function push(head_ref , new_data) {
var new_node = new Node();
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
function printList(node) {
while (node != null) {
document.write(" " + node.data);
node = node.next;
}
}
var a = null;
var b = null;
var intersect = null;
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
intersect = sortedIntersect(a, b);
document.write("\n Linked list containing "
+ "common items of a & b <br/> ");
printList(intersect);
</script>
|
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(max(m, n)).
The output list can store at most m+n nodes.
Method 4: Use Hashing
C++14
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node* next;
};
void printList(struct Node* node)
{
while (node != NULL) {
cout << " " << node->data;
node = node->next;
}
}
void append(struct Node** head_ref, int new_data)
{
struct Node* new_node
= (struct Node*)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
vector<int> intersection(struct Node* tmp1, struct Node* tmp2,
int k)
{
vector<int> res(k);
unordered_set<int> set;
while (tmp1 != NULL) {
set.insert(tmp1->data);
tmp1 = tmp1->next;
}
int cnt = 0;
while (tmp2 != NULL) {
if (set.find(tmp2->data) != set.end()) {
res[cnt] = tmp2->data;
cnt++;
}
tmp2 = tmp2->next;
}
return res;
}
int main()
{
struct Node* ll = NULL;
struct Node* ll1 = NULL;
append(&ll,7);
append(&ll,6);
append(&ll,5);
append(&ll,4);
append(&ll,3);
append(&ll,2);
append(&ll,1);
append(&ll,0);
append(&ll1,7);
append(&ll1,6);
append(&ll1,5);
append(&ll1,4);
append(&ll1,3);
append(&ll1,12);
append(&ll1,0);
append(&ll1,9);
vector<int> arr= intersection(ll, ll1, 6);
for (int i :arr)
cout << i << "\n";
return 0;
}
|
Java
import java.io.*;
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<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;
}
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);
}
}
}
|
Python3
class Node:
def __init__(self):
self.data = 0
self.next = None
def printList(node):
while (node != None):
print(node.data, end=" ")
node = node.next;
def append(head_ref, new_data):
new_node = Node()
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
def intersection(tmp1,tmp2,k):
res = [0]*k
set1 = set()
while (tmp1 != None):
set1.add(tmp1.data)
tmp1 = tmp1.next
cnt = 0
while (tmp2 != None):
if tmp2.data in set1:
res[cnt] = tmp2.data;
cnt += 1
tmp2 = tmp2.next
return res
def printList(node):
while (node != None):
print(node.data, end=" ")
node = node.next;
ll = None
ll1 = None
ll = append(ll , 7)
ll = append(ll , 6)
ll = append(ll , 5)
ll = append(ll , 4)
ll = append(ll , 3)
ll = append(ll , 2)
ll = append(ll , 1)
ll = append(ll , 0)
ll1 = append(ll1 , 7)
ll1 = append(ll1 , 6)
ll1 = append(ll1 , 5)
ll1 = append(ll1 , 4)
ll1 = append(ll1 , 3)
ll1 = append(ll1 , 12)
ll1 = append(ll1 , 0)
ll1 = append(ll1 , 9)
arr = intersection(ll , ll1 , 6)
for i in range(6):
print(arr[i])
|
C#
using System;
using System.Collections.Generic;
public class List {
Node head;
public class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
public void printList()
{
Node n = head;
while (n != null) {
Console.WriteLine(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<int> set = new HashSet<int>();
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;
}
public static void Main(String[] args)
{
List ll = new List();
List ll1 = new List();
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);
foreach(int i in arr) { Console.WriteLine(i); }
}
}
|
Javascript
class Node{
constructor(data){
this.data = data;
this.next = null;
}
}
function printList(node){
while(node != null){
document.write(node.data + " ");
node = node.next;
}
}
function append(head_ref, new_data){
new_node = new Node();
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
function intersection(tmp1, tmp2, k){
let res = new Array(k);
let set = new Set();
while(tmp1 != null){
set.add(tmp1.data);
tmp1 = tmp1.next;
}
let cnt = 0;
while(tmp2 != null){
if(set.has(tmp2.data)){
res[cnt] = tmp2.data;
cnt++;
}
tmp2 = tmp2.next;
}
return res;
}
let ll = null;
let ll1 = null;
ll = append(ll , 7)
ll = append(ll , 6)
ll = append(ll , 5)
ll = append(ll , 4)
ll = append(ll , 3)
ll = append(ll , 2)
ll = append(ll , 1)
ll = append(ll , 0)
ll1 = append(ll1 , 7)
ll1 = append(ll1 , 6)
ll1 = append(ll1 , 5)
ll1 = append(ll1 , 4)
ll1 = append(ll1 , 3)
ll1 = append(ll1 , 12)
ll1 = append(ll1 , 0)
ll1 = append(ll1 , 9)
let arr = intersection(ll, ll1, 6);
for(let i = 0; i<6; i++){
document.write(arr[i] + " ");
}
|
Complexity Analysis:
- Time Complexity: O(n)
- Space complexity: O(n) since auxiliary space is being used
Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.
References:
cslibrary.stanford.edu/105/LinkedListProblems.pdf
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...