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<stdio.h> #include<stdlib.h> #include<stdbool.h> // 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) { // Initialize current struct 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 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)? printf ( "Yes" ) : printf ( "No" ); return 0; } |
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<stdio.h> #include<stdlib.h> #include<stdbool.h> // 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)? printf ( "Yes" ) : printf ( "No" ); return 0; } |
Output:
Yes
Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for 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!
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