Overview
Merging two sorted linked lists into a single sorted linked list is a common problem in computer science, particularly in the context of linked list interview questions. This task is pivotal because it tests a candidate's understanding of linked lists, their manipulation, and the efficiency of different algorithms to merge sorted structures. It's a foundational problem for understanding more complex data structure manipulation and algorithm design.
Key Concepts
- Understanding of Linked Lists: Basic operations like traversal, insertion, and deletion.
- Algorithm Efficiency: Choosing the most efficient algorithm to merge two sorted lists, considering time and space complexity.
- Pointer Manipulation: Skills in handling node pointers to efficiently merge lists without creating a new list.
Common Interview Questions
Basic Level
- Explain how you would traverse a linked list.
- Write a function to append a node to a linked list.
Intermediate Level
- Describe the process of merging two sorted linked lists into a new sorted linked list.
Advanced Level
- How can you merge two sorted linked lists in place, without creating a new list, and what is its time complexity?
Detailed Answers
1. Explain how you would traverse a linked list.
Answer: To traverse a linked list, you start from the head node and continue visiting each node until you reach the end of the list (null). During traversal, you can perform operations such as searching for a value or printing each node's data.
Key Points:
- Start from the head of the list.
- Use a loop to visit each node until the next node is null.
- Access or modify the data of each node as needed during the traversal.
Example:
public class ListNode {
public int val;
public ListNode next;
public ListNode(int val=0, ListNode next=null) {
this.val = val;
this.next = next;
}
}
public void TraverseLinkedList(ListNode head) {
ListNode current = head;
while (current != null) {
Console.WriteLine(current.val);
current = current.next;
}
}
2. Write a function to append a node to a linked list.
Answer: Appending a node to a linked list requires you to traverse the list until you reach the last node, then set the next pointer of the last node to the new node.
Key Points:
- Traverse to the end of the list.
- Modify the next pointer of the last node to point to the new node.
- Consider the edge case where the list is initially empty.
Example:
public void AppendNode(ListNode head, int newValue) {
ListNode newNode = new ListNode(newValue);
if (head == null) {
head = newNode;
return;
}
ListNode current = head;
while (current.next != null) {
current = current.next;
}
current.next = newNode;
}
3. Describe the process of merging two sorted linked lists into a new sorted linked list.
Answer: The process involves creating a dummy head for the new sorted list and then comparing the nodes of the two lists one by one, appending the smaller value node to the new list until all nodes from both lists are added. The key is to iteratively compare the head nodes of both lists and link the smaller one to the new list, moving the pointer of the involved list one step forward.
Key Points:
- Use of a dummy head to simplify edge cases.
- Iterative comparison to maintain sorted order.
- Efficiently advancing the pointer of the list from which the node was taken.
Example:
public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
ListNode tail = dummy;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
tail.next = l1;
l1 = l1.next;
} else {
tail.next = l2;
l2 = l2.next;
}
tail = tail.next;
}
tail.next = l1 ?? l2;
return dummy.next;
}
4. How can you merge two sorted linked lists in place, without creating a new list, and what is its time complexity?
Answer: Merging two sorted linked lists in place involves adjusting the next pointers of the nodes in the lists so that they form a single, sorted list. This technique eliminates the need for a dummy head and additional space for a new list. By comparing the head nodes of both lists and rearranging the next pointers, you can merge the lists in place. The time complexity of this operation is O(n + m), where n and m are the lengths of the two lists, as each node in both lists is visited exactly once.
Key Points:
- In-place merging requires careful pointer manipulation.
- No additional space is needed for the list nodes.
- Time complexity is linear, O(n + m).
Example:
public ListNode MergeListsInPlace(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;
if (l1.val > l2.val) {
ListNode temp = l1;
l1 = l2;
l2 = temp;
}
ListNode res = l1;
while (l1 != null && l2 != null) {
ListNode tmp = null;
while (l1 != null && l1.val <= l2.val) {
tmp = l1;
l1 = l1.next;
}
tmp.next = l2;
// Swap
ListNode temp = l1;
l1 = l2;
l2 = temp;
}
return res;
}
By carefully manipulating pointers and swapping nodes as needed, we can efficiently merge two sorted linked lists in place without the need for additional space, making it a highly efficient operation.
