143. Reorder List LeetCode Solution

In this guide, you will get 143. Reorder List LeetCode Solution with the best time and space complexity. The solution to Reorder List problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Reorder List solution in C++
4. Reorder List solution in Java
5. Reorder List solution in Python
6. Additional Resources

Problem Statement of Reorder List

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:

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

Example 2:

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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(1)

143. Reorder List LeetCode 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 != nullptr && fast->next != nullptr) {
      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 != nullptr) {
      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;
    }
  }
};
/* code provided by PROGIEZ */

143. Reorder List LeetCode 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;
    }
  }
}
// code provided by PROGIEZ

143. Reorder List LeetCode 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)
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

Happy Coding! Keep following PROGIEZ for more updates and solutions.