面试题：从未排序的链表中删除重复项

Question

I'm reading Cracking the Coding Interview, Fourth Edition: 150 Programming Interview Questions and Solutions and I'm trying to solve the following question:

2.1 Write code to remove duplicates from an unsorted linked list. FOLLOW UP: How would you solve this problem if a temporary buffer is not allowed?

I'm solving it in C#, so I made my own Node class:

public class Node<T> where T : class
{
    public Node<T> Next { get; set; }
    public T Value { get; set; }

    public Node(T value)
    {
        Next = null;
        Value = value;
    }
}

My solution is to iterate through the list, then for each node to iterated through the remainder of the list and remove any duplicates (note that I haven't actually compiled or tested this, as instructed by the book):

public void RemoveDuplicates(Node<T> head)
{
    // Iterate through the list
    Node<T> iter = head;
    while(iter != null)
    {
        // Iterate to the remaining nodes in the list
        Node<T> current = iter;
        while(current!= null && current.Next != null)
        {
            if(iter.Value == current.Next.Value)
            {
                current.Next = current.Next.Next;
            }

            current = current.Next;
        }    

        iter = iter.Next;
    }
}

Here is the solution from the book (the author wrote it in java):

Without a buffer, we can iterate with two pointers: “current” does a normal iteration, while “runner” iterates through all prior nodes to check for dups. Runner will only see one dup per node, because if there were multiple duplicates they would have been removed already.

public static void deleteDups2(LinkedListNode head) 
{
    if (head == null) return;

    LinkedListNode previous = head;
    LinkedListNode current = previous.next;

    while (current != null) 
    {
        LinkedListNode runner = head;

        while (runner != current) { // Check for earlier dups
            if (runner.data == current.data) 
            {
                LinkedListNode tmp = current.next; // remove current
                previous.next = tmp;
                current = tmp; // update current to next node
                break; // all other dups have already been removed
            }
            runner = runner.next;
        }
        if (runner == current) { // current not updated - update now
            previous = current;
            current = current.next;
        }
    }
}

So my solution always looks for duplicates for the current node to the end, while their solution looks for duplicates from the head to the current node. I feel like both solutions would suffer performance issues depending on how many duplicates there are in the list and how they're distributed (density and position). But in general: is my answer nearly as good as the one in the book or is it significantly worse?

Answer 1

If you give a person a fish, they eat for a day. If you teach a person to fish...

My measures for the quality of an implementation are:

• Correctness : If you aren't getting the right answer in all cases, then it isn't ready
• Readability/maintainability : Look at code repetition, understandable names, the number of lines of code per block/method (and the number of things each block does), and how difficult it is to trace the flow of your code. Look at any number of books focused on refactoring, programming best-practices, coding standards, etc, if you want more information on this.
• Theoretical performance (worst-case and ammortized): Big-O is a metric you can use. CPU and memory consumption should both be measured
• Complexity : Estimate how it would take an average professional programmer to implement (if they already know the algorithm). See if that is in line with how difficult the problem actually is

As for your implementation:

• Correctness : I suggest writing unit tests to determine this for yourself and/or debugging it (on paper) from start to finish with interesting sample/edge cases. Null, one item, two items, various numbers of duplicates, etc
• Readability/maintainability : It looks mostly fine, though your last two comments don't add anything. It is a bit more obvious what your code does than the code in the book
• Performance : I believe both are N-squared. Whether the amortized cost is lower on one or the other I'll let you figure out :)
• Time to implement : An average professional should be able to code this algorithm in their sleep, so looking good

Answer 2

There's not much of a difference. If I've done my math right your's is on average N/16 slower than the authors but pleanty of cases exist where your implementation will be faster.

Edit:

I'll call your implementation Y and the author's A

Both proposed solutions has O(N^2) as worst case and they both have a best case of O(N) when all elements are the same value.

EDIT: This is a complete rewrite. Inspired by the debat in the comments I tried to find the average case for random N random numbers. That is a sequence with a random size and a random distribution. What would the average case be.

Y will always run U times where U is the number of unique numbers. For each iteration it will do NX comparisons where X is the number of elements removed prior to the iteration (+1). The first time no element will have been removed and on average on the second iteration N/U will have been removed.

That is on average ½N will been left to iterate. We can express the average cost as U*½N. The average U can be expressed based on N as well 0

Expressing A becomes more difficult. Let's say we use I iterations before we've encountered all unique values. After that will run between 1 and U comparisons (on average that's U/") and will do that NI times.

I*c+U/2(NI)

but whats the average number of comparisons (c) we run for the first I iterations. on average we need to compare against half of the elements already visited and on average we've visited I/2 elements, Ie. c=I/4

I/4+U/2(NI).

I can be expressed in terms of N. On average we'll need to visited half on N to find the unique values so I=N/2 yielding an average of

(I^2)/4+U/2(NI) which can be reduced to (3*N^2)/16.

That is of course if my estimation of the averages are correct. That is on average for any potential sequence A has N/16 fewer comparisons than Y but pleanty of cases exists where Y is faster than A. So I'd say they are equal when compared to the number of comparisons

Answer 3

How about using a HashMap? This way it will take O(n) time and O(n) space. I will write psuedocode.

function removeDup(LinkedList list){
  HashMap map = new HashMap();
  for(i=0; i<list.length;i++)
      if list.get(i) not in map
        map.add(list.get(i))
      else
        list.remove(i)
      end
  end
end

Of course we assume that HashMap has O(1) read and write.

Another solution is to use a mergesort and removes duplicate from start to end of the list. This takes O(n log n)

mergesort is O(n log n) removing duplicate from a sorted list is O(n). do you know why? therefore the entire operation takes O(n log n)

Answer 4

Here's the implementation using HashSet in O(n) time.

I have used a hashset to store unique values and 2 node-pointers to traverse through the linkedlist. If a duplicate is found, assign the value of current pointer to the previous pointer.

This will ensure removal of duplicate records.

    /// <summary>
    /// Write code to remove duplicates from an unsorted linked list.
    /// </summary>
    class RemoveDups<T>
    {
        private class Node
        {
            public Node Next;
            public T Data;
            public Node(T value)
            {
                this.Data = value;
            }
        }

        private Node head = null;

        public static void MainMethod()
        {
            RemoveDups<int> rd = new RemoveDups<int>();
            rd.AddNode(15);
            rd.AddNode(10);
            rd.AddNode(15);
            rd.AddNode(10);
            rd.AddNode(10);
            rd.AddNode(20);
            rd.AddNode(30);
            rd.AddNode(20);
            rd.AddNode(30);
            rd.AddNode(35);

            rd.PrintNodes();
            rd.RemoveDuplicates();

            Console.WriteLine("Duplicates Removed!");
            rd.PrintNodes();
        }

        private void RemoveDuplicates()
        {
            //use a hashtable to remove duplicates
            HashSet<T> hs = new HashSet<T>();
            Node current = head;
            Node prev = null;

            //loop through the linked list
            while (current != null)
            {
                if (hs.Contains(current.Data))
                {
                    //remove the duplicate record
                    prev.Next = current.Next;
                }
                else
                {
                    //insert element into hashset
                    hs.Add(current.Data);
                    prev = current;
                }
                current = current.Next;

            }
        }

        /// <summary>
        /// Add Node at the beginning
        /// </summary>
        /// <param name="val"></param>
        public void AddNode(T val)
        {
            Node newNode = new Node(val);
            newNode.Data = val;
            newNode.Next = head;
            head = newNode;
        }

        /// <summary>
        /// Print nodes
        /// </summary>
        public void PrintNodes()
        {
            Node current = head;
            while (current != null)
            {
                Console.WriteLine(current.Data);
                current = current.Next;
            }
        }
    }

Answer 5

Heapsort is an in-place sort. You could modify the "siftUp" or "siftDown" function to simply remove the element if it encounters a parent that is equal. This would be O(n log n)

function siftUp(a, start, end) is
 input:  start represents the limit of how far up the heap to sift.
               end is the node to sift up.
 child := end 
 while child > start
     parent := floor((child - 1) ÷ 2)
     if a[parent] < a[child] then (out of max-heap order)
         swap(a[parent], a[child])
         child := parent (repeat to continue sifting up the parent now)
     else if a[parent] == a[child] then
         remove a[parent]
     else
         return

Answer 6

Code in java:

public static void dedup(Node head) {
    Node cur = null;
    HashSet encountered = new HashSet();

    while (head != null) {
        encountered.add(head.data);
        cur = head;
        while (cur.next != null) {
            if (encountered.contains(cur.next.data)) {
                cur.next = cur.next.next;
            } else {
                break;
            }
        }
        head = cur.next;
    }
}

Answer 7

Tried the same in cpp. Please let me know your comments on this.

// ConsoleApplication2.cpp : Defines the entry point for the console application. //

#include "stdafx.h"
#include <stdlib.h>
struct node
{
    int data;
    struct node *next;
};
struct node *head = (node*)malloc(sizeof(node));
struct node *tail = (node*)malloc(sizeof(node));

struct node* createNode(int data)
{
    struct node *newNode = (node*)malloc(sizeof(node));
    newNode->data = data;
    newNode->next = NULL;
    head = newNode;
    return newNode;
}

bool insertAfter(node * list, int data)
{
    //case 1 - insert after head
    struct node *newNode = (node*)malloc(sizeof(node));
    if (!list)
    {

        newNode->data = data;
        newNode->next = head;
        head = newNode;
        return true;
    }

    struct node * curpos = (node *)malloc(sizeof(node));
    curpos = head;
    //case 2- middle, tail of list
    while (curpos)
    {
        if (curpos == list)
        {
            newNode->data = data;
            if (curpos->next == NULL)
            {
            newNode->next = NULL;
            tail = newNode;
            }
            else
            {
                newNode->next = curpos->next;
            }
            curpos->next = newNode;
            return true;
        }
        curpos = curpos->next;
    }
}

void deleteNode(node *runner, node * curr){

    //DELETE AT TAIL
    if (runner->next->next == NULL)
    {
        runner->next = NULL;        
    }
    else//delete at middle
    {
        runner = runner->next->next;
        curr->next = runner;
    }
    }


void removedups(node * list)
{
    struct node * curr = (node*)malloc(sizeof(node));
    struct node * runner = (node*)malloc(sizeof(node));
    curr = head;
    runner = curr;
    while (curr != NULL){
        runner = curr;
        while (runner->next != NULL){
            if (curr->data == runner->next->data){
                deleteNode(runner, curr);
            }
            if (runner->next!=NULL)
            runner = runner->next;
        }
        curr = curr->next;
    }
}
int _tmain(int argc, _TCHAR* argv[])
{
    struct node * list = (node*) malloc(sizeof(node));
    list = createNode(1);
    insertAfter(list,2);
    insertAfter(list, 2);
    insertAfter(list, 3);   
    removedups(list);
    return 0;
}

Answer 8

Code in C:

    void removeduplicates(N **r)
    {
        N *temp1=*r;
        N *temp2=NULL;
        N *temp3=NULL;
        while(temp1->next!=NULL)
        {
            temp2=temp1;
            while(temp2!=NULL)
            {
                temp3=temp2;
                temp2=temp2->next;
                if(temp2==NULL)
                {
                    break;
                }
                if((temp2->data)==(temp1->data))
                {
                    temp3->next=temp2->next;
                    free(temp2);
                    temp2=temp3;
                    printf("\na dup deleted");
                }
            }
            temp1=temp1->next;
        }

    }

Answer 9

Here's the answer in C

    void removeduplicates(N **r)
    {
        N *temp1=*r;
        N *temp2=NULL;
        N *temp3=NULL;
        while(temp1->next!=NULL)
        {
            temp2=temp1;
            while(temp2!=NULL)
            {
                temp3=temp2;
                temp2=temp2->next;
                if(temp2==NULL)
                {
                    break;
                }
                if((temp2->data)==(temp1->data))
                {
                    temp3->next=temp2->next;
                    free(temp2);
                    temp2=temp3;
                    printf("\na dup deleted");
                }
            }
            temp1=temp1->next;
        }

    }

Answer 10

您的解决方案与作者一样好，只有它在实现中有错误:)尝试在具有相同数据的两个节点的列表上进行跟踪。

Answer 11

Your approach is simply specular to the book's! You go forward, the book goes backward. There is no difference as both of you scan all elements. And, yes, since no buffer is allowed, there are performance issues. You usually don't have to mind about performance with such costrained questions and when it's not explicitly required.

Interview questions are made to test your open mindness. I have doubts about Mark's answer: it definitely is the best solution in real-world examples, but even if these algorithms use constant space, the constraint that no temporary buffer is allowed must be respected.

Otherwise, I guess that the book would have adopted such an approach. Mark, please forgive me for being critic against you.

Anyway, just to go deeper in the matter, yours and the book's approach both require Theta(n^2) time, while Mark's approach requires Theta(n logn) + Theta(n) time, which results in Theta(n logn) . Why Theta ? Because compare-swap algorithms are Omega(n logn) too, remember!

Answer 12

C# Code for removing duplicates left after the first set of iteration:

 public Node removeDuplicates(Node head) 
    {
        if (head == null)
            return head;

        var current = head;
        while (current != null)
        {
            if (current.next != null && current.data == current.next.data)
            {
                current.next = current.next.next;
            }
            else { current = current.next; }
        }

        return head;
    }

Answer 13

Hacker Rank Day24:More Linked Lists,Removing duplicate Node in C#.

static Node RemoveDuplicateNode(Node head)
    {
        Node Link = head;
        Node Previous;
        Node DulicateNode;
        int count = 0,temp;
        while (Link != null)
        {
            temp = Link.data;
            DulicateNode = Link;
            Previous = Link;
            while(DulicateNode != null)
            {
                if(DulicateNode.data==temp)
                {
                    Previous.data = DulicateNode.data;
                    Previous.next = DulicateNode.next;
                    ++count;
                }
                if(count>=2)
                {
                   if(DulicateNode.next != null)
                    {
                        DulicateNode.data = DulicateNode.next.data;
                        DulicateNode.next = DulicateNode.next.next;
                    }
                   else
                        DulicateNode=null;
                }
                else
                DulicateNode = DulicateNode.next;
            }
            count = 0;
            Link = Link.next;
        }
    
    
        return head;
    }