Introduction

In this blog post, we will explore how to solve the LeetCode problem "82" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well. Example 1: Input: head = [1,2,3,3,4,4,5] Output: [1,2,5] Example 2: Input: head = [1,1,1,2,3] Output: [2,3] Constraints: The number of nodes in the list is in the range [0, 300]. -100 <= Node.val <= 100 The list is guaranteed to be sorted in ascending order.

Explanation

Here's the breakdown of the solution:

• Iterate and Identify Duplicates: Traverse the linked list, identifying consecutive nodes with the same value.
• Skip Duplicates: When duplicates are found, advance the pointer until a distinct value is encountered, effectively removing the duplicate range.

• Maintain List Structure: Use a dummy head to simplify handling the beginning of the list and to properly link the remaining distinct nodes.

• Time Complexity: O(n), where n is the number of nodes in the linked list.

• Space Complexity: O(1)

Code

    class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def deleteDuplicates(head):
    """
    Given the head of a sorted linked list, delete all nodes that have duplicate
    numbers, leaving only distinct numbers from the original list. Return the
    linked list sorted as well.
    """
    dummy = ListNode(0)  # Dummy node to handle edge cases at the beginning
    dummy.next = head
    prev = dummy
    curr = head

    while curr:
        if curr.next and curr.val == curr.next.val:
            # Found duplicates, skip all of them
            while curr.next and curr.val == curr.next.val:
                curr = curr.next
            prev.next = curr.next  # Skip the entire duplicate sequence
        else:
            prev = curr  # No duplicates, move prev forward
        curr = curr.next  # Move curr forward

    return dummy.next