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Code Doubly linked list

Sort the biotonic doubly linked list

In this article, we will delve into the implementation of sorting a biotonic doubly linked list. A biotonic doubly linked list is a list that first increases up to a certain point and then decreases. The article will guide you through the process of sorting such a list while maintaining the biotonic property using Java code examples and detailed explanations.

Problem Statement

Given a biotonic doubly linked list, the task is to sort it while preserving its biotonic nature.


Suppose we have the following biotonic doubly linked list:

3 <-> 5 <-> 7 <-> 12 <-> 10 <-> 6 <-> 4 <-> 1 <-> -1

After sorting, the list should become:

-1 <-> 1 <-> 3 <-> 4 <-> 5 <-> 6 <-> 7 <-> 10 <-> 12

Idea to Solve

The idea is to traverse the linked list and identify the point where the decreasing order begins. Then, we can sort the increasing and decreasing segments separately, and finally, merge them to obtain the sorted biotonic list.


function sortBiotonicList():
    if head is not null:
        initialize front as head and last as null
        if <
            swap head and tail
            adjust pointers
        while front is not null and tail is not null and front is not tail:
            if <
                move last to the front of front node
                adjust pointers
                move front to the next node

Algorithm Explanation

  1. Implement a sortBiotonicList function.
  2. If the head is not null, initialize front with head and last as null.
  3. If is less than, swap head and tail, and adjust the pointers to maintain the doubly linked list structure.
  4. Traverse the list using a loop while front and tail are not null, and front is not equal to tail.
  5. If is less than, move the last node to the front of the front node. Adjust the pointers accordingly.
  6. Otherwise, move front to the next node.
  7. The list is now sorted in a biotonic manner.

Program Solution

Time Complexity

The time complexity of sorting a biotonic doubly linked list is O(n), where n is the number of nodes in the list. The list is traversed only once to identify the transition point and then sorted segments are merged.


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