Last Updated: 10 Dec, 2020
Search in a Linked List
Easy
Asked in company
You are given a Singly Linked List of integers with a head pointer. Every node of the Linked List has a value written on it.
A sample Linked List:
Now you have been given an integer value, 'K'. Your task is to check whether a node with a value equal to 'K' exists in the given linked list. Return 1 if node exists else return 0.
Input Format:
The first line of the input contains space-separated integers denoting the values of nodes of the Linked List.
The Linked List is terminated with -1. Hence, -1 is never a node value of the Linked List.
The second line contains a single integer 'K', which is desired to be checked in the Linked List.
Output Format:
The only line contains 1 if the desired value 'K' exists in the Linked List, otherwise, print 0.
Note:
You do not need to print anything; it has already been handled. Just implement the given function.
Approaches
Let's try to build a recursive solution to this problem
The recursive function has three cases:
- The Head is NULL(it means that we have reached the end of the linked list without finding value ‘K’ that means the particular value is not present in the Linked List. So, in this case, we will return 0 as our answer).
- The Head is Not NULL, and the value of the head is equal to the desired value ‘K’ we will return 1 in this case as we found the desired value in the Linked List.
- The Head is Not NULL, and the current value is not equal to the desired value ‘K’, so it will recur to the next node of the head and undergo the same process there.
The idea is to first initialize a pointer pointing to the first element of the Linked List and will check if the pointer's data is equal to the desired value ‘K’. If yes, then we will return 1; otherwise, we will move the pointer ahead by 1 position. If at any point of time during our travel, the pointer gets to point the NULL node, i.e., the end of Linked List, we will return 0 (as we have reached the end).