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
Tree: Level Order Traversal
Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F
View Solution →Binary Search Tree : Insertion
You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <
View Solution →Tree: Huffman Decoding
Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t
View Solution →Binary Search Tree : Lowest Common Ancestor
You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b
View Solution →Swap Nodes [Algo]
A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from
View Solution →Kitty's Calculations on a Tree
Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a
View Solution →