Sort List LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Sort List in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either Easy, Medium, or Hard.

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• Mathematics/Basic Logical Based Questions
• Arrays
• Strings
• Hash Table
• Dynamic Programming
• Stack & Queue
• Trees & Graphs
• Greedy Algorithms
• Breadth-First Search
• Depth-First Search
• Sorting & Searching
• BST (Binary Search Tree)
• Database
• Linked List
• Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the Problem – Sort List– LeetCode Problem

Sort List– LeetCode Problem

Problem:

Given the head of a linked list, return the list after sorting it in ascending order.

Example 1:

Input: head = [4,2,1,3]
Output: [1,2,3,4]

Example 2:

Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]

Example 3:

Input: head = []
Output: []

Constraints:

• The number of nodes in the list is in the range [0, 5 * 104].
• -105 <= Node.val <= 105

Sort List– LeetCode Solutions

Sort List Solution in C++:

class Solution {
 public:
  ListNode* sortList(ListNode* head) {
    const int length = getLength(head);
    ListNode dummy(0, head);

    for (int k = 1; k < length; k *= 2) {
      ListNode* curr = dummy.next;
      ListNode* tail = &dummy;
      while (curr) {
        ListNode* l = curr;
        ListNode* r = split(l, k);
        curr = split(r, k);
        auto [mergedHead, mergedTail] = merge(l, r);
        tail->next = mergedHead;
        tail = mergedTail;
      }
    }

    return dummy.next;
  }

 private:
  int getLength(ListNode* head) {
    int length = 0;
    for (ListNode* curr = head; curr; curr = curr->next)
      ++length;
    return length;
  }

  ListNode* split(ListNode* head, int k) {
    while (--k && head)
      head = head->next;
    ListNode* rest = head ? head->next : nullptr;
    if (head)
      head->next = nullptr;
    return rest;
  }

  pair<ListNode*, ListNode*> merge(ListNode* l1, ListNode* l2) {
    ListNode dummy(0);
    ListNode* tail = &dummy;

    while (l1 && l2) {
      if (l1->val > l2->val)
        swap(l1, l2);
      tail->next = l1;
      l1 = l1->next;
      tail = tail->next;
    }
    tail->next = l1 ? l1 : l2;
    while (tail->next)
      tail = tail->next;

    return {dummy.next, tail};
  }
};

Sort List Solution in Java:

class Solution {
  public ListNode sortList(ListNode head) {
    final int length = getLength(head);
    ListNode dummy = new ListNode(0, head);

    for (int k = 1; k < length; k *= 2) {
      ListNode curr = dummy.next;
      ListNode tail = dummy;
      while (curr != null) {
        ListNode l = curr;
        ListNode r = split(l, k);
        curr = split(r, k);
        ListNode[] merged = merge(l, r);
        tail.next = merged[0];
        tail = merged[1];
      }
    }

    return dummy.next;
  }

  private int getLength(ListNode head) {
    int length = 0;
    for (ListNode curr = head; curr != null; curr = curr.next)
      ++length;
    return length;
  }

  private ListNode split(ListNode head, int k) {
    while (--k > 0 && head != null)
      head = head.next;
    ListNode rest = head == null ? null : head.next;
    if (head != null)
      head.next = null;
    return rest;
  }

  private ListNode[] merge(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode tail = dummy;

    while (l1 != null && l2 != null) {
      if (l1.val > l2.val) {
        ListNode temp = l1;
        l1 = l2;
        l2 = temp;
      }
      tail.next = l1;
      l1 = l1.next;
      tail = tail.next;
    }
    tail.next = l1 == null ? l2 : l1;
    while (tail.next != null)
      tail = tail.next;

    return new ListNode[] {dummy.next, tail};
  }
}

Sort List Solution in Python:

class Solution:
  def sortList(self, head: ListNode) -> ListNode:
    def split(head: ListNode, k: int) -> ListNode:
      while k > 1 and head:
        head = head.next
        k -= 1
      rest = head.next if head else None
      if head:
        head.next = None
      return rest

    def merge(l1: ListNode, l2: ListNode) -> tuple:
      dummy = ListNode(0)
      tail = dummy

      while l1 and l2:
        if l1.val > l2.val:
          l1, l2 = l2, l1
        tail.next = l1
        l1 = l1.next
        tail = tail.next
      tail.next = l1 if l1 else l2
      while tail.next:
        tail = tail.next

      return dummy.next, tail

    length = 0
    curr = head
    while curr:
      length += 1
      curr = curr.next

    dummy = ListNode(0, head)

    k = 1
    while k < length:
      curr = dummy.next
      tail = dummy
      while curr:
        l = curr
        r = split(l, k)
        curr = split(r, k)
        mergedHead, mergedTail = merge(l, r)
        tail.next = mergedHead
        tail = mergedTail
      k *= 2

    return dummy.next