Overview
Merging two sorted linked lists into a single sorted linked list is a common problem in computer science, often encountered in scenarios where data from different sources needs to be combined and processed in a sorted manner. This operation is crucial for maintaining sorted sequences in data structures and algorithms and is widely used in interview questions to assess understanding of linked lists, pointers, and recursion.
Key Concepts
- Pointer Manipulation: Understanding how to manipulate pointers or references to nodes within linked lists.
- Merge Algorithm: Knowing how to merge two lists while maintaining their sort order.
- Recursion vs Iteration: Comparing recursive and iterative approaches to solving the merge problem.
Common Interview Questions
Basic Level
- How do you merge two sorted linked lists into a single sorted linked list using an iterative approach?
- Explain the recursive approach to merge two sorted linked lists.
Intermediate Level
- How would you handle merging when the linked lists have different lengths?
Advanced Level
- Can you optimize the space complexity of merging two sorted linked lists?
Detailed Answers
1. How do you merge two sorted linked lists into a single sorted linked list using an iterative approach?
Answer: To merge two sorted linked lists iteratively, you maintain a pointer to the last node of the result list and advance the pointers in the input lists to choose the smaller current element at each step. Initialize a dummy head to simplify edge cases and use it to keep track of the start of the merged list.
Key Points:
- Start with a dummy node to simplify edge cases and maintain a tail pointer for the last node of the merged list.
- Compare the current nodes of both lists, attaching the smaller one to the tail of the merged list, and advance the pointer in that list.
- Continue until you reach the end of one of the lists, then attach the remainder of the other list to the merged list.
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 ListNode MergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
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;
}
if (l1 != null) {
tail.next = l1;
} else if (l2 != null) {
tail.next = l2;
}
return dummy.next;
}
2. Explain the recursive approach to merge two sorted linked lists.
Answer: The recursive approach involves breaking down the problem into smaller subproblems. If both lists are not empty, compare the first elements of each list. Recursively merge the rest of the lists, then set the next pointer of the smaller element to the result of the recursive call.
Key Points:
- Base cases are when either of the lists is null, in which case the non-null list (or null) is returned as the result.
- The recursion goes deeper by calling itself with the next node of the list that had the smaller first element and the other list unchanged.
- Recursive calls build the merged list backwards, connecting smaller elements at each step.
Example:
public ListNode MergeTwoListsRecursive(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;
if (l1.val < l2.val) {
l1.next = MergeTwoListsRecursive(l1.next, l2);
return l1;
} else {
l2.next = MergeTwoListsRecursive(l1, l2.next);
return l2;
}
}
3. How would you handle merging when the linked lists have different lengths?
Answer: Both the iterative and recursive approaches inherently handle lists of different lengths. After the end of one list is reached, the remaining nodes of the other list are still in sorted order and can be directly appended to the result list. This ensures that the merged list is complete and sorted, regardless of the initial lengths of the input lists.
Key Points:
- The process of comparing nodes continues until one list is exhausted.
- The remainder of the longer list is already sorted and can be appended directly.
- No special handling is required beyond the logic in the basic merge algorithm.
Example: The examples provided in questions 1 and 2 already cater to this scenario as they continue the merge operation until both lists are fully traversed.
4. Can you optimize the space complexity of merging two sorted linked lists?
Answer: The iterative and recursive approaches already have an optimal space complexity of O(1) and O(n) respectively, where n is the depth of the recursion stack. The iterative approach does not use extra space for additional nodes, operating directly on the input lists. The recursive approach, however, uses stack space for recursive calls. To optimize, you could avoid recursion in favor of iteration to eliminate the stack space usage.
Key Points:
- The iterative approach is already space-optimized as it does not allocate additional nodes.
- The recursive approach's space complexity depends on the depth of recursion, which corresponds to the length of the lists.
- Converting a recursive approach to an iterative one can help in achieving space optimization in scenarios where stack space is a concern.
Example: Refer to the iterative method provided in question 1 for an example of a space-optimized approach.
