Get Node Value
Problem Statement :
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the tail. The value 2 is at the desired position. Complete the getNode function in the editor below. getNode has the following parameters: SinglyLinkedListNode pointer head: refers to the head of the list int positionFromTail: the item to retrieve Returns int: the value at the desired position Input Format The first line contains an integer t, the number of test cases. Each test case has the following format: The first line contains an integer n, the number of elements in the linked list. The next n lines contains an integer, the data value for an element of the linked list. The last line contains an integer positionFromTail, the position from the tail to retrieve the value of. Function Description
Solution :
Solution in C :
In C++ :
/*
Get Nth element from the end in a linked list of integers
Number of elements in the list will always be greater than N.
Node is defined as
struct Node
{
int data;
struct Node *next;
}
*/
int GetNode(Node *head,int positionFromTail)
{
// This is a "method-only" submission.
// You only need to complete this method.
int count = 0;
Node *ptr = head;
while(ptr!=NULL)
{
++count;
ptr=ptr->next;
}
count = count-positionFromTail;
ptr=head;
--count;
while(count!=0)
{
--count;
ptr=ptr->next;
}
return ptr->data;
}
In python3 :
def GetNode(head, position):
nums = []
current = head
# First find tail of list as length is unknown
while current != None:
nums.append(current.data)
current = current.next
# Return value of item at given position
return nums[ len( nums ) - position - 1 ]
In java :
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
}
*/
int GetNode(Node head,int n) {
// This is a "method-only" submission.
// You only need to complete this method.
if(head.next==null)
return head.data;
Node temp=head;
int count=0;
while(temp!=null)
{
count++;
temp=temp.next;
}
int k=count-n-1;
temp=head;
while(k>0)
{
k--;
temp=temp.next;
}
return temp.data;
}
In C :
// Complete the getNode function below.
/*
* For your reference:
*
* SinglyLinkedListNode {
* int data;
* SinglyLinkedListNode* next;
* };
*
*/
int getNode(SinglyLinkedListNode* head, int positionFromTail) {
int c = 1, i = 1;
SinglyLinkedListNode *ptr = head;
while(ptr->next != NULL){
c += 1;
ptr = ptr->next;
}
if(positionFromTail == 0)
return ptr->data;
ptr = head;
c = c - positionFromTail;
while(i != c){
ptr = ptr->next;
i += 1;
}
return ptr->data;
}
View More Similar Problems
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 →Tree : Top View
Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.
View Solution →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 →