Linked List Interview Questions and Answers
Linked list is a data structure that provides a way of storing and organizing data, similar to a dynamic array. However, unlike an array which stores data in an indexed manner, a linked list stores data in an interconnected manner. In a linked list, each piece of data is stored in a node, which contains the data and a reference to the next node in the list. Here are some of Linked List Interview Questions and Answers.
What is a linked list and how is it different from an array?
A linked list is a data structure consisting of a sequence of nodes that each store a piece of data and a reference to the next node in the list. This data structure provides a way of organizing data that is different from an array, which stores data in an indexed manner.
The primary difference between a linked list and an array is that in a linked list, each node contains a reference to the next node in the list, while in an array, each element points to a single index. This means that linked list nodes can be dynamically added or removed from the list without having to shift the remaining elements around. This dynamic nature makes linked lists more efficient than arrays in some cases, as adding or removing elements from an array can be time-consuming.
In a linked list, each node can store any type of data, while in an array, all elements must be the same type. Additionally, linked lists are typically unordered, meaning that the order of elements cannot be determined without traversing the list, while arrays are naturally ordered.
How do you traverse a linked list ?
Traversing a linked list involves visiting each node in the list and performing a specific operation on it. The most basic way to traverse a linked list is to use a loop, such as a for loop, to iterate through the list and perform an operation on each node. For example, to print the contents of a linked list, you could use the following code:
//Given a linked list head
Node* head = //head of linked list;
//Traverse list
Node* current = head;
while (current != nullptr) {
std::cout << current->data;
current = current->next;
}
In this code, we start by assigning the head of the list to the “current” pointer. Then, we use a while loop to traverse the list and print the contents of each node. Each time the loop iterates, we set the “current” pointer to the next node in the list, until we reach the end of the list.
What operations can be performed on a linked list?
Linked lists can be manipulated using a variety of operations. Common operations include adding and removing elements from the list, searching for an element, sorting the list, and reversing the list.
Adding elements to a linked list is done by creating a new node, connecting it to the list, and updating the head pointer if necessary. Removing elements from a list is done by unlinking the node from the list and updating the head pointer if necessary.
Searching for an element in a linked list is done by comparing each node’s data to the data being searched for. Sorting a linked list is done by sorting the underlying data and then relinking the nodes in the sorted order. Reversing a linked list is done by iterating through the list and relinking each node to its predecessor.
How do you reverse a linked list?
It is one of the most frequently asked Linked List Interview Questions.
Reversing a linked list involves iterating through the list and relinking each node to its predecessor. This can be done using a loop and a temporary pointer to keep track of the previous node. For example, to reverse a linked list with the following structure:
A -> B -> C -> D -> E
We could use the following code:
//Given a linked list head
Node* head = //head of linked list;
Node* current = head;
Node* prev = nullptr;
//Traverse list
while (current != nullptr) {
Node* temp = current->next;
current->next = prev;
prev = current;
current = temp;
}
//Set head to new head
head = prev;
In this code, we start by setting the “current” pointer to the head of the list, and the “prev” pointer to null. Then, we use a while loop to iterate through the list and relink each node to its predecessor. Each time the loop iterates, we keep track of the next node in the list using a temporary pointer and then relink the current node to its predecessor. Finally, we set the head pointer to the new head of the list.
How do you find the middle element of a linked list?
Finding the middle element of a linked list involves using two pointers, one that travels twice as fast as the other. This is done by having one pointer iterate through the list, while the other pointer is set to the beginning of the list. As the first pointer iterates, the second pointer will eventually reach the middle element of the list.
For example, to find the middle element of a linked list with the following structure:
A -> B -> C -> D -> E -> F
We could use the following code:
//Given a linked list head
Node* head = //head of linked list;
Node* slow = head;
Node* fast = head;
//Traverse list
while (fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
}
//Middle element is at slow pointer
std::cout << slow->data;
In this code, we start by setting the “slow” and “fast” pointers to the head of the list. Then, we use a while loop to iterate through the list. Each time the loop iterates, we increment the “slow” pointer by one and the “fast” pointer by two. When the “fast” pointer reaches the end of the list, the “slow” pointer will be at the middle element of the list. Finally, we print the data at the “slow” pointer.
Hope you are familiar with the famous Linked List Interview Questions and got its answers as well. Thanks for reading!!!
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