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用于从排序链表中删除重复项的 C#程序
用于从排序链表中删除重复项的 C# 程序
原文:https://www . geesforgeks . org/c-program-for-remove-replications-from-sorted-link-list-2/
编写一个函数，接受按非递减顺序排序的列表，并从列表中删除任何重复的节点。该列表只能遍历一次。 例如，如果链接列表是 11->11->11->21->43->43->60，则 removeDuplicates()应该将列表转换为 11- > 21- > 43- > 60。
算法: 从头(或开始)节点遍历列表。遍历时，将每个节点与其下一个节点进行比较。如果下一个节点的数据与当前节点相同，则删除下一个节点。在删除节点之前，我们需要存储节点的下一个指针
实现: removeDuplicates()之外的函数只是创建一个链表并测试 remove duplicates()。
C
// C# program to remove duplicates
// from a sorted linked list
using System;
public class LinkedList
{
// Head of list
Node head;
// Linked list Node
class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
void removeDuplicates()
{
// Another reference to head
Node current = head;
/* Pointer to store the next
pointer of a node to be deleted*/
Node next_next;
// Do nothing if the list is empty
if (head == null)
return;
// Traverse list till the last node
while (current.next != null)
{
// Compare current node with the
// next node
if (current.data == current.next.data)
{
next_next = current.next.next;
current.next = null;
current.next = next_next;
}
// Advance if no deletion
else
current = current.next;
}
}
// Utility functions
// Inserts a new Node at front
// of the list.
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
// 3\. Make next of new Node as head
new_node.next = head;
// 4\. Move the head to point to new Node
head = new_node;
}
// Function to print linked list
void printList()
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
// Driver code
public static void Main(String []args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(13);
llist.push(13);
llist.push(11);
llist.push(11);
llist.push(11);
Console.WriteLine(
"List before removal of duplicates");
llist.printList();
llist.removeDuplicates();
Console.WriteLine(
"List after removal of elements");
llist.printList();
}
}
// This code is contributed by 29AjayKumar
输出:
Linked list before duplicate removal 11 11 11 13 13 20
Linked list after duplicate removal 11 13 20
时间复杂度: O(n)，其中 n 是给定链表中的节点数。
递归方法:
C
// C# Program to remove duplicates
// from a sorted linked list
using System;
class GFG
{
// Link list node
public class Node
{
public int data;
public Node next;
};
// The function removes duplicates
// from a sorted list
static Node removeDuplicates(Node head)
{
/* Pointer to store the pointer
of a node to be deleted*/
Node to_free;
// Do nothing if the list is empty
if (head == null)
return null;
// Traverse the list till last node
if (head.next != null)
{
// Compare head node with next node
if (head.data == head.next.data)
{
/* The sequence of steps is important.
to_free pointer stores the next of
head pointer which is to be deleted.*/
to_free = head.next;
head.next = head.next.next;
removeDuplicates(head);
}
// This is tricky: only advance if no deletion
else
{
removeDuplicates(head.next);
}
}
return head;
}
// UTILITY FUNCTIONS
/* Function to insert a node at the
beginning of the linked list */
static Node push(Node head_ref,
int new_data)
{
// Allocate node
Node new_node = new Node();
// Put in the data
new_node.data = new_data;
// Link the old list off the new node
new_node.next = (head_ref);
// Move the head to point to the
// new node
(head_ref) = new_node;
return head_ref;
}
/* Function to print nodes in
a given linked list */
static void printList(Node node)
{
while (node != null)
{
Console.Write(" " + node.data);
node = node.next;
}
}
// Driver code
public static void Main(String []args)
{
// Start with the empty list
Node head = null;
/* Let us create a sorted linked list
to test the functions. Created
linked list will be
11.11.11.13.13.20 */
head = push(head, 20);
head = push(head, 13);
head = push(head, 13);
head = push(head, 11);
head = push(head, 11);
head = push(head, 11);
Console.Write("Linked list before" +
" duplicate removal ");
printList(head);
// Remove duplicates from linked list
head = removeDuplicates(head);
Console.Write("Linked list after" +
" duplicate removal ");
printList(head);
}
}
// This code is contributed by PrinciRaj1992
输出:
Linked list before duplicate removal 11 11 11 13 13 20
Linked list after duplicate removal 11 13 20
另一种方法:创建一个指向每个元素第一次出现的指针和另一个将迭代到每个元素的指针 temp，当前一个指针的值不等于 temp 指针时，我们将前一个指针的指针设置为另一个节点的第一次出现。
下面是上述方法的实现:
C
// C# program to remove duplicates
// from a sorted linked list
using System;
class LinkedList
{
// Head of list
public Node head;
// Linked list Node
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
// Function to remove duplicates
// from the given linked list
void removeDuplicates()
{
// Two references to head temp
// will iterate to the whole
// Linked List prev will point
// towards the first occurrence
// of every element
Node temp = head, prev = head;
// Traverse list till the last node
while (temp != null)
{
// Compare values of both pointers
if(temp.data != prev.data)
{
/* if the value of prev is not
equal to the value of temp
that means there are no more
occurrences of the prev data.
So we can set the next of
prev to the temp node.*/
prev.next = temp;
prev = temp;
}
/*Set the temp to the next node*/
temp = temp.next;
}
/* This is the edge case if there are more
than one occurrences of the last element */
if(prev != temp)
{
prev.next = null;
}
}
// Utility functions
// Inserts a new Node at front
// of the list.
public void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
// 3\. Make next of new Node as head
new_node.next = head;
// 4\. Move the head to point to
// new Node
head = new_node;
}
// Function to print linked list
void printList()
{
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
// Driver code
public static void Main(string []args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(13);
llist.push(13);
llist.push(11);
llist.push(11);
llist.push(11);
Console.Write("List before ");
Console.WriteLine("removal of duplicates");
llist.printList();
llist.removeDuplicates();
Console.WriteLine(
"List after removal of elements");
llist.printList();
}
}
// This code is contributed by rutvik_56
输出:
List before removal of duplicates
11 11 11 13 13 20
List after removal of elements
11 13 20
另一种方法:使用地图
这个想法是推送地图中的所有值并打印其关键点。
下面是上述方法的实现:
C
// C# program for the above approach
using System;
using System.Collections.Generic;
public class Node
{
public int data;
public Node next;
public Node()
{
data = 0;
next = null;
}
}
public class GFG
{
/* Function to insert a node at
the beginning of the linked list */
static Node push(Node head_ref,
int new_data)
{
// Allocate node
Node new_node = new Node();
// Put in the data
new_node.data = new_data;
// Link the old list off
// the new node
new_node.next = (head_ref);
/* Move the head to point
to the new node */
head_ref = new_node;
return head_ref;
}
/* Function to print nodes
in a given linked list */
static void printList(Node node)
{
while (node != null)
{
Console.Write(node.data + " ");
node = node.next;
}
}
// Function to remove duplicates
static void removeDuplicates(Node head)
{
Dictionary<int, bool> track =
new Dictionary<int, bool>();
Node temp = head;
while(temp != null)
{
if(!track.ContainsKey(temp.data))
{
Console.Write(temp.data + " ");
track.Add(temp.data , true);
}
temp = temp.next;
}
}
// Driver Code
static public void Main ()
{
Node head = null;
/* Created linked list will be
11->11->11->13->13->20 */
head = push(head, 20);
head = push(head, 13);
head = push(head, 13);
head = push(head, 11);
head = push(head, 11);
head = push(head, 11);
Console.Write(
"Linked list before duplicate removal ");
printList(head);
Console.Write(
"Linked list after duplicate removal ");
removeDuplicates(head);
}
}
// This code is contributed by rag2127
输出:
Linked list before duplicate removal 11 11 11 13 13 20
时间复杂度: O(节点数)
空间复杂度: O(节点数)
更多详情请参考从排序链表中删除重复项的完整文章！
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