算法设计:从未排序的链表中删除重复项

2021年3月10日16:17:41 发表评论 761 次浏览

本文概述

编写一个removeDuplicates()函数, 该函数获取一个列表并从列表中删除所有重复的节点。该列表未排序。

例如, 如果链接列表为12-> 11-> 12-> 21-> 41-> 43-> 21, 则removeDuplicates()会将列表转换为12-> 11-> 21-> 41-> 43。

推荐:请在"

实践

首先, 在继续解决方案之前。

方法1(使用两个循环)

这是使用两个循环的简单方法。外循环用于一个接一个地拾取元素, 内循环将拾取的元素与其余元素进行比较。

感谢Gaurav Saxena在编写此代码方面的帮助。

C ++

/* Program to remove duplicates in an unsorted
    linked list */
#include<bits/stdc++.h>
using namespace std;
  
/* A linked list node */
struct Node
{
     int data;
     struct Node *next;
};
  
// Utility function to create a new Node
struct Node *newNode( int data)
{
    Node *temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}
  
/* Function to remove duplicates from a
    unsorted linked list */
void removeDuplicates( struct Node *start)
{
     struct Node *ptr1, *ptr2, *dup;
     ptr1 = start;
  
     /* Pick elements one by one */
     while (ptr1 != NULL && ptr1->next != NULL)
     {
         ptr2 = ptr1;
  
         /* Compare the picked element with rest
            of the elements */
         while (ptr2->next != NULL)
         {
             /* If duplicate then delete it */
             if (ptr1->data == ptr2->next->data)
             {
                 /* sequence of steps is important here */
                 dup = ptr2->next;
                 ptr2->next = ptr2->next->next;
                 delete (dup);
             }
             else /* This is tricky */
                 ptr2 = ptr2->next;
         }
         ptr1 = ptr1->next;
     }
}
  
/* Function to print nodes in a given linked list */
void printList( struct Node *node)
{
     while (node != NULL)
     {
         printf ( "%d " , node->data);
         node = node->next;
     }
}
  
/* Druver program to test above function */
int main()
{
     /* The constructed linked list is:
      10->12->11->11->12->11->10*/
     struct Node *start = newNode(10);
     start->next = newNode(12);
     start->next->next = newNode(11);
     start->next->next->next = newNode(11);
     start->next->next->next->next = newNode(12);
     start->next->next->next->next->next =
                                     newNode(11);
     start->next->next->next->next->next->next =
                                     newNode(10);
  
     printf ( "Linked list before removing duplicates " );
     printList(start);
  
     removeDuplicates(start);
  
     printf ( "\nLinked list after removing duplicates " );
     printList(start);
  
     return 0;
}

Java

// Java program to remove duplicates from unsorted 
// linked list
  
class LinkedList {
  
     static Node head;
  
     static class Node {
  
         int data;
         Node next;
  
         Node( int d) {
             data = d;
             next = null ;
         }
     }
  
     /* Function to remove duplicates from an
        unsorted linked list */
     void remove_duplicates() {
         Node ptr1 = null , ptr2 = null , dup = null ;
         ptr1 = head;
  
         /* Pick elements one by one */
         while (ptr1 != null && ptr1.next != null ) {
             ptr2 = ptr1;
  
             /* Compare the picked element with rest
                 of the elements */
             while (ptr2.next != null ) {
  
                 /* If duplicate then delete it */
                 if (ptr1.data == ptr2.next.data) {
  
                     /* sequence of steps is important here */
                     dup = ptr2.next;
                     ptr2.next = ptr2.next.next;
                     System.gc();
                 } else /* This is tricky */ {
                     ptr2 = ptr2.next;
                 }
             }
             ptr1 = ptr1.next;
         }
     }
  
     void printList(Node node) {
         while (node != null ) {
             System.out.print(node.data + " " );
             node = node.next;
         }
     }
  
     public static void main(String[] args) {
         LinkedList list = new LinkedList();
         list.head = new Node( 10 );
         list.head.next = new Node( 12 );
         list.head.next.next = new Node( 11 );
         list.head.next.next.next = new Node( 11 );
         list.head.next.next.next.next = new Node( 12 );
         list.head.next.next.next.next.next = new Node( 11 );
         list.head.next.next.next.next.next.next = new Node( 10 );
  
         System.out.println( "Linked List before removing duplicates : \n " );
         list.printList(head);
  
         list.remove_duplicates();
         System.out.println( "" );
         System.out.println( "Linked List after removing duplicates : \n " );
         list.printList(head);
     }
}
// This code has been contributed by Mayank Jaiswal

C#

// C# program to remove duplicates from unsorted 
// linked list
using System;
class List_{
  
   Node head;
  
   class Node{
  
     public
       int data;
     public
       Node next;
  
     public
  
     Node( int d) 
     {
       data = d;
       next = null ;
     }
   }
  
   /* Function to remove duplicates from an
        unsorted linked list */
   void remove_duplicates() 
   {
     Node ptr1 = null , ptr2 = null , dup = null ;
     ptr1 = head;
  
     /* Pick elements one by one */
     while (ptr1 != null && 
            ptr1.next != null )
     {
       ptr2 = ptr1;
  
       /* Compare the picked element with rest
                 of the elements */
       while (ptr2.next != null ) 
       {
  
         /* If duplicate then delete it */
         if (ptr1.data == ptr2.next.data) 
         {
  
           /* sequence of steps is important here */
           dup = ptr2.next;
           ptr2.next = ptr2.next.next;
  
         }
         else /* This is tricky */ 
         {
           ptr2 = ptr2.next;
         }
       }
       ptr1 = ptr1.next;
     }
   }
  
   void printList(Node node)
   {
     while (node != null ) 
     {
       Console.Write(node.data + " " );
       node = node.next;
     }
   }
  
   // Driver Code
   public static void Main(String[] args) 
   {
     List_ list = new List_();
     list.head = new Node(10);
     list.head.next = new Node(12);
     list.head.next.next = new Node(11);
     list.head.next.next.next = new Node(11);
     list.head.next.next.next.next = new Node(12);
     list.head.next.next.next.next.next = new Node(11);
     list.head.next.next.next.next.next.next = new Node(10);
  
     Console.WriteLine( "Linked List_ before removing duplicates : \n " );
     list.printList(list.head);
  
     list.remove_duplicates();
     Console.WriteLine( "" );
     Console.WriteLine( "Linked List_ after removing duplicates : \n " );
     list.printList(list.head);
   }
}
  
// This code is contributed by gauravrajput1

输出: 

Linked list before removing duplicates:
 10 12 11 11 12 11 10 
Linked list after removing duplicates:
 10 12 11

时间复杂度:O(n ^ 2)

方法2(使用排序)

通常, 合并排序是最适合对链表进行有效排序的排序算法

1)使用合并排序对元素进行排序。我们很快将写一篇有关对链表进行排序的文章。 O(nLogn)

2)使用

删除已排序链表中重复项的算法。上)

请注意, 此方法不会保留元素的原始顺序。

时间复杂度:O(nLogn)

方法3(使用散列)

我们从头到尾遍历链接列表。对于每个新遇到的元素, 我们检查它是否在哈希表中:如果是, 则将其删除;否则, 将其删除。否则我们将其放在哈希表中。

C ++

/* Program to remove duplicates in an unsorted
    linked list */
#include<bits/stdc++.h>
using namespace std;
  
/* A linked list node */
struct Node
{
     int data;
     struct Node *next;
};
  
// Utility function to create a new Node
struct Node *newNode( int data)
{
    Node *temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}
  
/* Function to remove duplicates from a
    unsorted linked list */
void removeDuplicates( struct Node *start)
{
     // Hash to store seen values
     unordered_set< int > seen;
  
     /* Pick elements one by one */
     struct Node *curr = start;
     struct Node *prev = NULL;
     while (curr != NULL)
     {
         // If current value is seen before
         if (seen.find(curr->data) != seen.end())
         {
            prev->next = curr->next;
            delete (curr);
         }
         else
         {
            seen.insert(curr->data);
            prev = curr;
         }
         curr = prev->next;
     }
}
  
/* Function to print nodes in a given linked list */
void printList( struct Node *node)
{
     while (node != NULL)
     {
         printf ( "%d " , node->data);
         node = node->next;
     }
}
  
/* Driver program to test above function */
int main()
{
     /* The constructed linked list is:
      10->12->11->11->12->11->10*/
     struct Node *start = newNode(10);
     start->next = newNode(12);
     start->next->next = newNode(11);
     start->next->next->next = newNode(11);
     start->next->next->next->next = newNode(12);
     start->next->next->next->next->next =
                                     newNode(11);
     start->next->next->next->next->next->next =
                                     newNode(10);
  
     printf ( "Linked list before removing duplicates : \n" );
     printList(start);
  
     removeDuplicates(start);
  
     printf ( "\nLinked list after removing duplicates : \n" );
     printList(start);
  
     return 0;
}

Java

// Java program to remove duplicates
// from unsorted linkedlist
  
import java.util.HashSet;
  
public class removeDuplicates 
{
     static class node 
     {
         int val;
         node next;
  
         public node( int val) 
         {
             this .val = val;
         }
     }
      
     /* Function to remove duplicates from a
        unsorted linked list */
     static void removeDuplicate(node head) 
     {
         // Hash to store seen values
         HashSet<Integer> hs = new HashSet<>();
      
         /* Pick elements one by one */
         node current = head;
         node prev = null ;
         while (current != null ) 
         {
             int curval = current.val;
              
              // If current value is seen before
             if (hs.contains(curval)) {
                 prev.next = current.next;
             } else {
                 hs.add(curval);
                 prev = current;
             }
             current = current.next;
         }
  
     }
      
     /* Function to print nodes in a given linked list */
     static void printList(node head) 
     {
         while (head != null ) 
         {
             System.out.print(head.val + " " );
             head = head.next;
         }
     }
  
     public static void main(String[] args) 
     {
         /* The constructed linked list is:
          10->12->11->11->12->11->10*/
         node start = new node( 10 );
         start.next = new node( 12 );
         start.next.next = new node( 11 );
         start.next.next.next = new node( 11 );
         start.next.next.next.next = new node( 12 );
         start.next.next.next.next.next = new node( 11 );
         start.next.next.next.next.next.next = new node( 10 );
  
         System.out.println( "Linked list before removing duplicates :" );
         printList(start);
  
         removeDuplicate(start);
  
         System.out.println( "\nLinked list after removing duplicates :" );
         printList(start);
     }
}
  
// This code is contributed by Rishabh Mahrsee

C#

// C# program to remove duplicates
// from unsorted linkedlist
using System;
using System.Collections.Generic;
  
class removeDuplicates{
class node 
{
     public int val;
     public node next;
  
     public node( int val) 
     {
         this .val = val;
     }
}
  
// Function to remove duplicates from a
// unsorted linked list 
static void removeDuplicate(node head) 
{
      
     // Hash to store seen values
     HashSet< int > hs = new HashSet< int >();
  
     // Pick elements one by one 
     node current = head;
     node prev = null ;
     while (current != null ) 
     {
         int curval = current.val;
          
         // If current value is seen before
         if (hs.Contains(curval))
         {
             prev.next = current.next;
         }
         else 
         {
             hs.Add(curval);
             prev = current;
         }
         current = current.next;
     }
}
  
// Function to print nodes in a 
// given linked list 
static void printList(node head) 
{
     while (head != null ) 
     {
         Console.Write(head.val + " " );
         head = head.next;
     }
}
  
// Driver code
public static void Main(String[] args) 
{
      
     // The constructed linked list is:
     // 10->12->11->11->12->11->10
     node start = new node(10);
     start.next = new node(12);
     start.next.next = new node(11);
     start.next.next.next = new node(11);
     start.next.next.next.next = new node(12);
     start.next.next.next.next.next = new node(11);
     start.next.next.next.next.next.next = new node(10);
  
     Console.WriteLine( "Linked list before removing " +
                       "duplicates :" );
     printList(start);
  
     removeDuplicate(start);
  
     Console.WriteLine( "\nLinked list after removing " + 
                       "duplicates :" );
     printList(start);
}
}
  
// This code is contributed by amal kumar choubey

输出:

Linked list before removing duplicates:
 10 12 11 11 12 11 10 
Linked list after removing duplicates:
 10 12 11

感谢bearwang建议这种方法。

时间复杂度:平均O(n)(假设哈希表访问时间平均为O(1))。

如果你发现以上任何解释/算法不正确, 或者是解决同一问题的更好方法, 请写评论。

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