710293. Sort a k sorted doubly linked list
Sort a k sorted doubly linked list
Slug: sort-a-k-sorted-doubly-linked-list
Difficulty: Medium
Id: 710293
Topic Tags: Linked List, Data Structures
Company Tags: None
Summary
Given a doubly linked list where each node contains at most k nodes in sorted order, the task is to sort the entire list. This problem involves manipulating a linked list and can be solved using various algorithms.
Detailed Explanation
To solve this problem, we'll use a simple merge sort approach. Since we have k-sorted lists, we can take advantage of this property by merging adjacent lists in sorted order. We start by splitting the input list into smaller segments of size k each. Each segment is already sorted, so we can simply merge these segments together to form the final sorted list.
Here's a step-by-step breakdown of the solution:
- Split the input list into segments of size
k. This will give us an array of linked lists, where each linked list contains at mostknodes. - For each segment, we'll create a new node that contains the smallest element from all the linked lists in that segment. We can do this by iterating through the linked lists and keeping track of the minimum value seen so far.
- Add the new node to our result list and remove it from its original position in the segment.
- Repeat steps 2-3 until we've processed all segments.
- The final sorted list will be the concatenation of the nodes in each segment, with the smallest element first.
Time complexity: O(n log k) where n is the total number of nodes and k is the maximum size of a segment.
Space complexity: O(n) since we need to store the result list.
Optimized Solutions
Java
public Node sortKSortedDoublyLinkedList(Node head, int k) {
List<Node> segments = new ArrayList<>();
Node current = head;
while (current != null) {
Node next = current.next;
if (next == null || current.data.compareTo(next.data) > 0) {
// Start a new segment
segments.add(current);
current = next;
} else {
current = next;
}
}
Node result = new Node();
while (!segments.isEmpty()) {
Node minNode = segments.get(0);
for (int i = 1; i < segments.size(); i++) {
if (minNode.data.compareTo(segments.get(i).data) > 0) {
minNode = segments.get(i);
}
}
result.next = minNode;
segments.remove(minNode);
}
return result.next;
}