Remove Duplicates from Unsorted List

Given the head of a unsorted linked list, delete all duplicates such that each element appears only once. Return the linked list as well.

Solutions

Approach 1 – Using hash table

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Easy solution, use hash table to check duplicate item.

Steps

1. iterate through the linked list and adding each element to a hash table.

2. Whenwe discover a duplicate element, we remove the element and continue iterating.

3. Return head node.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode DeleteDuplicates(ListNode head)
    {
        HashSet<int> set = new HashSet<int>();
        ListNode previous = null;
        if (head == null) return null;

        ListNode temp = head;
        while (temp != null)
        {
            if (set.Contains(temp.val))
            {
                previous.next = temp.next;
            }
            else
            {
                set.Add(temp.val);
                previous = temp;
            }


            temp = temp.next;
        }
        return previous;

    }
}

Time Complexity: O(n)

Auxiliary Space: O(n)

Approach 2 – No Buffer Allowed

---

lf we don't have a buffer, we can iterate with two pointers: current which iterates through the linked list, and runner which checks all subsequent nodes for duplicates.

Steps

1. iterate through the linked list and adding each element to a hash table.

2. Whenwe discover a duplicate element, we remove the element and continue iterating.

3. Return head node.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode DeleteDuplicates(ListNode head)
    {
        ListNode current = head;
        while (current != null)
        {
            ListNode runner = current;
            while (runner.next != null)
            {
                if (current.val == runner.next.val)
                {
                    runner.next = runner.next.next;
                }
                else
                {
                    runner = runner.next;
                }

            }
            current = current.next;
        }
        return head;

    }
}

Time Complexity: O(n²)

Auxiliary Space: O(1)