**Compare two linked lists**

### Problem Statement :

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lists are not equal. Return 0. Function Description: Complete the compare_lists function in the editor below. compare_lists has the following parameters: 1. SinglyLinkedListNode llist1: a reference to the head of a list 2. SinglyLinkedListNode llist2: a reference to the head of a list Returns: int: return 1 if the lists are equal, or 0 otherwise Input Format: The first line contains an integer t, the number of test cases. Each of the test cases has the following format: The first line contains an integer n, the number of nodes in the first linked list. Each of the next n lines contains an integer, each a value for a data attribute. The next line contains an integer m, the number of nodes in the second linked list. Each of the next m lines contains an integer, each a value for a data attribute. Constraints: 1. 1<=t<=10 2. 1<=n,m<=1000 3. 1<=list1[i], list2[i]<=1000

### Solution :

` ````
Solution in C :
In C:
//the following function is all that is needed to complete the
//challenge in hackerrank platform.
bool compare_lists(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
struct SinglyLinkedListNode *t1;
struct SinglyLinkedListNode *t2;
t1=head1;t2 = head2;
if(t1==NULL && t2==NULL)
return 1;
if(t1 != NULL && t2 == NULL)
return 0;
if(t1 == NULL && t2 != NULL)
return 0;
else
{
while(t1->next != NULL && t2->next != NULL)
{
if(t1->data == t2->data)
{
t1 = t1->next;
t2 = t2->next;
}
else return 0;
}
if(t1->next == NULL && t2->next == NULL)
return 1;
else return 0;
}
}
In C++:
//the following function is all that is needed to complete the
//challenge in hackerrank platform.
int CompareLists(Node *headA, Node* headB)
{
// This is a "method-only" submission.
// You only need to complete this method
int flag=0;
while(headA != NULL && headB != NULL)
{
if(headA->data!=headB->data)
{
flag=1;
break;
}
headA = headA->next;
headB = headB->next;
}
if(flag==1 || headA!=NULL || headB!=NULL)
return 0;
else
return 1;
}
In Java:
//the following method is all that is needed to complete the
//challenge in hackerrank platform.
static boolean compareLists(SinglyLinkedListNode head1, SinglyLinkedListNode head2) {
if (head1 == null && head2 == null) {
return true;
} else if (head1 == null || head2 == null) {
return false;
} else if (head1.data != head2.data) {
return false;
}
return compareLists(head1.next,head2.next);
}
In Python 3:
//the following method is all that is needed to complete the
//challenge in hackerrank platform.
def CompareLists(headA, headB):
if (headA==None) | (headB==None):
if (headA==None) & (headB==None):
return 1
else:
return 0
else:
while (headA!=None) & (headB!=None):
if (headA.data == headB.data):
headA = headA.next
headB = headB.next
else:
return 0
if (headA == None) & (headB == None):
return 1
else:
return 0
```

## View More Similar Problems

## Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

View Solution →## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

View Solution →## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

View Solution →## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

View Solution →## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →