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.
Java
// Iterative Java program to search // an element in linked list//Node classclass Node{ int data; Node next; Node(int d) { data = d; next = null; }}//Linked list classclass LinkedList{ // Head of list Node head; // Inserts a new node at the front // of the list public void push(int new_data) { //Allocate new node and putting data Node 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 public boolean search(Node head, int x) { // Initialize current Node current = head; while (current != null) { // Data found if (current.data == x) return true; current = current.next; } // Data not found return false; } // Driver code public static void main(String args[]) { // Start with the empty list LinkedList llist = new LinkedList(); // Use push() to construct list // 14->21->11->30->10 llist.push(10); llist.push(30); llist.push(11); llist.push(21); llist.push(14); if (llist.search(llist.head, 21)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Pratik Agarwal |
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.
Java
// Recursive Java program to search an element// in linked list// Node classclass Node{ int data; Node next; Node(int d) { data = d; next = null; }}// Linked list classclass LinkedList{ // Head of list Node head; // Inserts a new node at the // front of the list public void push(int new_data) { // Allocate new node and putting data Node 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 public boolean search(Node head, int 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 public static void main(String args[]) { // Start with the empty list LinkedList llist = new LinkedList(); // Use push() to construct list // 14->21->11->30->10 llist.push(10); llist.push(30); llist.push(11); llist.push(21); llist.push(14); if (llist.search(llist.head, 21)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by Pratik Agarwal |
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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