Merging multiple sorted linked lists into a single sorted linked list is a common challenge in coding interviews and real-world applications. By mastering this problem, you’ll gain valuable insights into techniques like merging algorithms, data structures, and optimization strategies. Let’s break it down step by step.
Understanding the Problem: Merge k Sorted Linked Lists
You are given an array of k sorted linked lists. The task is to merge all the linked lists into one sorted linked list and return its head.
Key Details
• Input: An array of k sorted linked lists.
• Output: A single sorted linked list.
• Constraints:
• k can be large.
• The total number of nodes across all linked lists is n .
Example:
• Input:
lists = [[1,4,5], [1,3,4], [2,6]]
• Output:
[1, 1, 2, 3, 4, 4, 5, 6]
Starting with the Naive Approach to Merge k Sorted Lists
When tackling the “Merge k Sorted Lists” problem, it’s essential to begin with the naive approach. This method provides a simple, intuitive way to merge multiple sorted linked lists one by one. While not the most efficient, it helps establish a foundational understanding of the problem before moving on to optimized solutions.
public class MergeKSortedLists {
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) return null;
ListNode merged = null;
for (ListNode list : lists) {
merged = mergeTwoLists(merged, list);
}
return merged;
}
// Helper function to merge two sorted linked lists
private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0); // Dummy node to simplify list construction
ListNode current = dummy;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
current.next = l1;
l1 = l1.next;
} else {
current.next = l2;
l2 = l2.next;
}
current = current.next;
}
// Append the remaining elements from either list
current.next = (l1 != null) ? l1 : l2;
return dummy.next;
}
}
The mergeKLists function solves the problem of merging multiple sorted lists by repeatedly using mergeTwoLists, which handles the merging of two sorted lists at a time. This iterative approach ensures that the final output is a single sorted linked list.
Why This Code Is Not Efficient
• What Happens:
• At each step, the mergeTwoLists function combines the merged list with another input list.
• This results in the merged list growing larger with each iteration, making subsequent merges increasingly expensive.
• Example:
• If k = 5 , you first merge lists L1 and L2 , then merge the result with L3 , and so on.
• By the time you merge the last list, the merged list contains almost all nodes, leading to repetitive comparisons.
Efficiently Merging k Sorted Lists Using Divide and Conquer
The naive approach to merging k sorted lists involves merging lists sequentially, one at a time. While simple, this results in a time complexity of O(k log n) , where n is the total number of nodes across all lists. This becomes inefficient as k grows larger.
The divide-and-conquer approach:
1. Divides the array of lists into smaller groups, reducing the problem size at each step.
2. Recursively Merges these smaller groups, combining them efficiently.
3. Combines Two Lists at a time using the mergeTwoLists helper function, which ensures sorted order.
This reduces the time complexity to O(n log k) , making it much more suitable for larger inputs.
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) return null;
return mergeKListsHelper(lists, 0, lists.length - 1);
}
private ListNode mergeKListsHelper(ListNode[] lists, int start, int end) {
if (start == end) return lists[start];
if (start < end) {
int mid = start + (end - start) / 2;
ListNode left = mergeKListsHelper(lists, start, mid);
ListNode right = mergeKListsHelper(lists, mid + 1, end);
return mergeTwoLists(left, right);
}
return null;
}
private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode current = dummy;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
current.next = l1;
l1 = l1.next;
} else {
current.next = l2;
l2 = l2.next;
}
current = current.next;
}
current.next = (l1 != null) ? l1 : l2;
return dummy.next;
}
}
Step-by-Step Explanation
1. Calculate the Midpoint
int mid = start + (end - start) / 2;• Why the Midpoint?:
• It ensures that the input is split evenly, creating balanced subproblems.
• Balanced subproblems minimize the depth of the recursion tree, leading to optimal performance.
2. Recursively Merge the Left & Right Half
ListNode left = mergeKListsHelper(lists, start, mid);
ListNode right = mergeKListsHelper(lists, mid + 1, end);
The left & right half is solved independently, reducing the problem size.
3. Merge
return mergeTwoLists(left, right);• Use the mergeTwoLists function to combine the two halves into a single sorted list.
Conclusion
Finding the midpoint is a cornerstone of the divide-and-conquer approach because it ensures:
• Balanced splitting of subproblems.
• Efficient recursion and merging.
• Optimal time complexity, making the algorithm scalable for large k . Without the midpoint, the problem becomes harder to manage and far less efficient.
