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 nodeclass 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 codeint 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 nodestruct 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 listbool 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 codeint 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!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
