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.
Javascript
<script> // Iterative javascript program // to search an element // in linked list //Node class class Node { constructor(d) { this .data = d; this .next = null ; } } // Linked list class // Head of list var head; // Inserts a new node at the front of the list function push(new_data) { // Allocate new node and putting data var new_node = new Node(new_data); // Make next of new node as head new_node.next = head; // Move the head to point to new Node head = new_node; } // Checks whether the value // x is present in linked list function search(head , x) { // Initialize current var current = head; while (current != null ) { if (current.data == x) // Data found return true ; current = current.next; } // Data not found return false ; } // Driver code // Start with the empty list // Use push() to construct list // 14->21->11->30->10 push(10); push(30); push(11); push(21); push(14); if (search(head, 21)) document.write( "Yes" ); else document.write( "No" ); // This code contributed by aashish1995 </script> |
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.
Javascript
<script> // Recursive javascript program to search // an element in linked list // Node class class Node { constructor(val) { this .data = val; this .next = null ; } } // Linked list class // Head of list var head; // Inserts a new node at the front // of the list function push(new_data) { // Allocate new node and putting data var new_node = new Node(new_data); // Make next of new node as head new_node.next = head; // Move the head to point to new Node head = new_node; } // Checks whether the value x is present // in linked list function search(head, x) { // Base case if (head == null ) return false ; // If key is present in current node, // return true if (head.data == x) return true ; // Recur for remaining list return search(head.next, x); } // Driver code // Start with the empty list // Use push() to construct list // 14->21->11->30->10 push(10); push(30); push(11); push(21); push(14); if (search(head, 21)) document.write( "Yes" ); else document.write( "No" ); // This code contributed by gauravrajput1 </script> |
Output:
Yes
Time Complexity: O(n), where n represents the length of the given linked list.
Auxiliary Space: O(n), for recursive 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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