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.
// 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.
// 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!