Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Search an element in the given singly Linked List using C++
Given a singly linked list, the task is to search a particular element in the linked list. If the element is found, then print “present” otherwise “Not present”. For example,
Input-1 −
1→ 2→ 3→ 4→ 5→ 6
Searching for ‘7’
Output −
Not Present
Explanation − In the given singly linked list the element ‘7’ is not present, hence we will return the output as “Not Present”.
Input-2 −
1→ 2→ 3→ 4→ 5
Searching for ‘2’
Output −
Present
Explanation − Since in the given Singly linked list the element ‘2’ is present thus we will return the output as “Present”.
Approach to solve this problem
There are two approaches to search a particular element in the given singly linked list; we have to check recursively whether one element is present in the linked list or not.
If the linked list is empty, we will return false otherwise if the current node having the data value is equal to the input element then we will return true. In the other approach, we check the element iteratively if it equals the current head pointer or not and return true or false accordingly.
Take Input and Initialize a singly linked list by inserting nodes in it.
A Boolean recursive function searhRecursive(node*head, int element) takes the head pointer of the linked list and the key element as a parameter.
Initially if the head is NULL or if the linked list is empty then return false.
If the element to be searched is equal to the current head of the linked list then return true.
Example
#include<iostream>
using namespace std;
#include<iostream>
using namespace std;
class node{
public:
int data;
node*next;
node(int d){
data=d;
node*next= NULL;
}
};
void insertAt(node*&head, int data){
node*n= new node(data);
n->next= head;
head= n;
}
bool searchRecursive(node*head,int key){
if(head==NULL){
return false;
}
if(head->data==key){
return true;
}
else{
return searchRecursive(head->next, key);
}
}
void printNode(node*head){
while(head!=NULL){
cout<<head->data<<"->";
head=head->next;
}
cout<<endl;
}
int main(){
node*head= NULL;
insertAt(head,5);
insertAt(head,4);
insertAt(head,3);
insertAt(head,2);
insertAt(head,1);
printNode(head);
if(searchRecursive(head,7)){
cout<<"present"<<endl;
}
else{
cout<<"Not Present"<<endl;
}
}Output
Running the above code will generate the output as,
Not Present
Since in the given linked list 1→2→3→4→5 the element ‘7’ is not present thus we return “Not Present”.
385 Views
