Reorder 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 Reorder 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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  • 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 – Reorder List– LeetCode Problem 

Reorder List– LeetCode Problem

Problem:

You are given the head of a singly linked-list. The list can be represented as:

L0 → L1 → … → Ln - 1 → Ln

Reorder the list to be on the following form:

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

You may not modify the values in the list’s nodes. Only nodes themselves may be changed.

Example 1:

reorder1linked list
Input: head = [1,2,3,4]
Output: [1,4,2,3]

Example 2:

reorder2 linked list
Input: head = [1,2,3,4,5]
Output: [1,5,2,4,3]

Constraints:

  • The number of nodes in the list is in the range [1, 5 * 104].
  • 1 <= Node.val <= 1000
Reorder List– LeetCode Solutions
Reorder List Solution in C++:
class Solution {
 public:
  void reorderList(ListNode* head) {
    if (!head || !head->next)
      return;

    ListNode* mid = findMid(head);
    ListNode* reversed = reverse(mid);
    merge(head, reversed);
  }

 private:
  ListNode* findMid(ListNode* head) {
    ListNode* prev = nullptr;
    ListNode* slow = head;
    ListNode* fast = head;

    while (fast && fast->next) {
      prev = slow;
      slow = slow->next;
      fast = fast->next->next;
    }
    prev->next = nullptr;

    return slow;
  }

  ListNode* reverse(ListNode* head) {
    ListNode* prev = nullptr;
    ListNode* curr = head;

    while (curr) {
      ListNode* next = curr->next;
      curr->next = prev;
      prev = curr;
      curr = next;
    }

    return prev;
  }

  void merge(ListNode* l1, ListNode* l2) {
    while (l2) {
      ListNode* next = l1->next;
      l1->next = l2;
      l1 = l2;
      l2 = next;
    }
  }
};
Reorder List Solution in Java:
class Solution {
  public void reorderList(ListNode head) {
    if (head == null || head.next == null)
      return;

    ListNode mid = findMid(head);
    ListNode reversed = reverse(mid);
    merge(head, reversed);
  }

  private ListNode findMid(ListNode head) {
    ListNode prev = null;
    ListNode slow = head;
    ListNode fast = head;

    while (fast != null && fast.next != null) {
      prev = slow;
      slow = slow.next;
      fast = fast.next.next;
    }
    prev.next = null;

    return slow;
  }

  private ListNode reverse(ListNode head) {
    ListNode prev = null;
    ListNode curr = head;

    while (curr != null) {
      ListNode next = curr.next;
      curr.next = prev;
      prev = curr;
      curr = next;
    }

    return prev;
  }

  private void merge(ListNode l1, ListNode l2) {
    while (l2 != null) {
      ListNode next = l1.next;
      l1.next = l2;
      l1 = l2;
      l2 = next;
    }
  }
}
Reorder List Solution in Python:
class Solution:
  def reorderList(self, head: ListNode) -> None:
    def findMid(head: ListNode):
      prev = None
      slow = head
      fast = head

      while fast and fast.next:
        prev = slow
        slow = slow.next
        fast = fast.next.next
      prev.next = None

      return slow

    def reverse(head: ListNode) -> ListNode:
      prev = None
      curr = head

      while curr:
        next = curr.next
        curr.next = prev
        prev = curr
        curr = next

      return prev

    def merge(l1: ListNode, l2: ListNode) -> None:
      while l2:
        next = l1.next
        l1.next = l2
        l1 = l2
        l2 = next

    if not head or not head.next:
      return

    mid = findMid(head)
    reversed = reverse(mid)
    merge(head, reversed)

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