C++ Program For Searching An Element In A Linked List

Write a function that searches a given key ‘x’ in a given singly linked list. The function should return true if x is present in linked list and false otherwise.

bool search(Node *head, int x)

For example, if the key to be searched is 15 and linked list is 14->21->11->30->10, then function should return false. If key to be searched is 14, then the function should return true.
Iterative Solution:

1) Initialize a node pointer, current = head.
2) Do following while current is not NULL
    a) current->key is equal to the key being searched return true.
    b) current = current->next
3) Return false 

Following is iterative implementation of above algorithm to search a given key. 

C++

// Iterative C++ program to search
// an element in linked list
#include <bits/stdc++.h>
using namespace std;

// Link list node
class Node
{
    public:
    int key;
    Node* next;
};

/* Given a reference (pointer to pointer) to
   the head of a list and an int, push a new
   node on the front of the list. */
void push(Node** head_ref, int new_key)
{
    // Allocate node
    Node* new_node = new Node();

    // Put in the key
    new_node->key = new_key;

    // Link the old list of the
    // new node
    new_node->next = (*head_ref);

    // Move the head to point to the
    // new node
    (*head_ref) = new_node;
}

// Checks whether the value x is
// present in linked list
bool search(Node* head, int x)
{
    Node* current = head;
    while (current != NULL)
    {
        if (current->key == x)
            return true;
        current = current->next;
    }
    return false;
}

// Driver code
int main()
{
    // Start with the empty list
    Node* head = NULL;
    int x = 21;

    // Use push() to construct list
    // 14->21->11->30->10
    push(&head, 10);
    push(&head, 30);
    push(&head, 11);
    push(&head, 21);
    push(&head, 14);

    search(head, 21)? cout<<"Yes" : cout<<"No";
    return 0;
}
// This is code is contributed by rathbhupendra

Output:

Yes

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Recursive Solution:

bool search(head, x)
1) If head is NULL, return false.
2) If head's key is same as x, return true;
3) Else return search(head->next, x) 

Following is the recursive implementation of the above algorithm to search a given key.

C++

// Recursive C++ program to search
// an element in linked list
#include <bits/stdc++.h>
using namespace std;

// Link list node
struct Node
{
    int key;
    struct Node* next;
};

/* Given a reference (pointer to pointer) to
   the head of a list and an int, push a new
   node on the front of the list. */
void push(struct Node** head_ref,
          int new_key)
{
    // Allocate node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));

    // Put in the key
    new_node->key = new_key;

    // Link the old list of the new node
    new_node->next = (*head_ref);

    // Move the head to point to
    // the new node
    (*head_ref) = new_node;
}

// Checks whether the value x is
// present in linked list
bool search(struct Node* head, int x)
{
    // Base case
    if (head == NULL)
        return false;

    // If key is present in current
    // node, return true
    if (head->key == x)
        return true;

    // Recur for remaining list
    return search(head->next, x);
}

// Driver code
int main()
{
    // Start with the empty list
    struct Node* head = NULL;
    int x = 21;

    // Use push() to construct list
    // 14->21->11->30->10
    push(&head, 10);
    push(&head, 30);
    push(&head, 11);
    push(&head, 21);
    push(&head, 14);

    search(head, 21)? cout << "Yes" : cout << "No";
    return 0;
}
// This code is contributed by SHUBHAMSINGH10

Output:

Yes

Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), due to recursive call stack where n represents the length of the given linked list.

Please refer complete article on Search an element in a Linked List (Iterative and Recursive) for more details!