使用递归从中间顺序到左右顺序遍历链表

2021年4月25日17:08:54 发表评论 955 次浏览

本文概述

给定一个链表。任务是使用递归从中间顺序到左右顺序遍历链表

例如:

如果给定的链表是:2-> 5-> 8-> 3-> 7-> 9-> 12-> NULL
中间从左到右的顺序是:3, 8, 7, 5, 5, 9, 2, 12

说明:给定链表的中间是'3', 所以我们从中间开始遍历, 先打印3, 然后在3的左边和右边, 所以我们打印8、7, 然后再打印8的左边, 然后在7的右边, 所以我们打印5、9, 然后在左边打印5, 在右边打印9, 因此我们打印2、12。

注意:如果链表中的节点数为偶数, 则仅从左向右打印。对于此链表(包含偶数个节点)2-> 5-> 8-> 7-> 9-> 12-> NULL。

输出应为8、7、5、9、2、12。

例子:

输入:20-> 15-> 23-> 13-> 19-> 32-> 16-> 41-> 11-> NULL
输出:19、13、32、23、16、15、41、20、11
输入:12-> 25-> 51-> 16-> 9-> 90-> 7-> 2-> NULL
输出:16, 9, 51, 90, 25, 7, 12, 2。

方法:

首先, 计算链表的大小:

  • 如果大小为奇数:
    ->然后使用递归转到第(n + 1)/ 2个节点。
  • 如果大小是偶数:
    ->然后使用递归转到第n / 2个节点。
  • 现在打印节点数据并返回下一个节点地址, 除非函数调用堆栈为空, 否则请执行此步骤。

下面是上述方法的实现:

C ++

//A C++ program to demonstrate
//the printing of Linked List middle
//to left right order
  
#include <bits/stdc++.h>
using namespace std;
  
//A linked list node
class Node {
public :
     int data;
     Node* next;
};
  
//Given a reference (pointer to pointer)
//to the head of a list and an int, appends
//a new node at the end
  
void append(Node** head_ref, int new_data)
{
     //Allocate node
     Node* new_node = new Node();
  
     //Used in step 5
     Node* last = *head_ref;
  
     //Put in the data
     new_node->data = new_data;
  
     //This new node is going to be
     //the last node, so make next of
     //it as NULL
     new_node->next = NULL;
  
     //If the Linked List is empty, //then make the new node as head
     if (*head_ref == NULL) {
         *head_ref = new_node;
         return ;
     }
  
     //Else traverse till the last node
     while (last->next != NULL)
         last = last->next;
  
     //Change the next of last node
     last->next = new_node;
     return ;
}
  
//This function prints contents of
//linked list starting from head
  
void printList(Node* node)
{
     while (node != NULL) {
         cout <<" " <<node->data;
  
         if (node->next != NULL)
             cout <<"->" ;
         node = node->next;
     }
}
  
//Function to get the size of linked list
int getSize(Node* head)
{
     if (head == NULL)
         return 0;
     return 1 + getSize(head->next);
}
  
//Utility function to print the Linked List
//from middle to left right order
Node* printMiddleToLeftRightUtil(Node* head, int counter, int lSize)
{
     //Base Condition
     //When size of list is odd
     if (counter == 1 && lSize % 2 != 0) {
  
         //Print node value
         cout <<head->data;
  
         //Returns address of next node
         return head->next;
     }
  
     //Base Condition
     //When size of list is even
     else if (counter == 1) {
  
         //Print node value
         //and next node value
         cout <<head->data;
         cout <<" , " <<head->next->data;
  
         //Returns address of next to next node
         return head->next->next;
     }
     else {
  
         //Recursive function call and
         //store return address
         Node* ptr = printMiddleToLeftRightUtil(
             head->next, counter - 1, lSize);
  
         //Print head data
         cout <<" , " <<head->data;
  
         //Print ptr data
         cout <<" , " <<ptr->data;
  
         //Returns address of next node
         return ptr->next;
     }
}
  
//Function to print Middle to
//Left-right order
void printMiddleToLeftRight(Node* head)
{
     //Function call to get the size
     //Of Linked List
     int listSize = getSize(head);
  
     int middle = 0;
  
     //Store middle when Linked
     //List size is odd
     if (listSize % 2 != 0) {
         middle = (listSize + 1) /2;
     }
  
     //Store middle when Linked
     //List size is even
  
     else {
         middle = listSize /2;
     }
  
     //Utility function call print
     //Linked List from Middle
     //to left right order
     cout <<"Output : " ;
  
     printMiddleToLeftRightUtil(head, middle, listSize);
}
  
//Driver code
int main()
{
     //Start with the empty list
     Node* head = NULL;
  
     //Insert 6. So linked list
     //becomes 6->NULL
     append(&head, 6);
  
     //Insert 6. So linked list
     //becomes 6->4->NULL
     append(&head, 4);
     append(&head, 8);
     append(&head, 7);
     append(&head, 9);
     append(&head, 11);
     append(&head, 2);
  
     //After inserting linked list
     //becomes 6->4->8->7->9->11->2->NULL
     cout <<"Created Linked list is: " ;
  
     //Function to display Linked List content
     printList(head);
     cout <<endl;
  
     //Function call print Linked List from
     //Middle to left right order
     printMiddleToLeftRight(head);
     return 0;
}

Java

//A Java program to demonstrate
//the printing of Linked List middle
//to left right order
class GFG
{
  
//A linked list node
static class Node
{
     int data;
     Node next;
};
  
//Given a reference (pointer to pointer)
//to the head of a list and an int, appends
//a new node at the end
static Node append(Node head_ref, int new_data)
{
     //Allocate node
     Node new_node = new Node();
  
     //Used in step 5
     Node last = head_ref;
  
     //Put in the data
     new_node.data = new_data;
  
     //This new node is going to be
     //the last node, so make next of
     //it as null
     new_node.next = null ;
  
     //If the Linked List is empty, //then make the new node as head
     if (head_ref == null )
     {
         head_ref = new_node;
         return head_ref;
     }
  
     //Else traverse till the last node
     while (last.next != null )
         last = last.next;
  
     //Change the next of last node
     last.next = new_node;
     return head_ref;
}
  
//This function prints contents of
//linked list starting from head
static void printList(Node node)
{
     while (node != null ) 
     {
         System.out.print( " " + node.data);
  
         if (node.next != null )
             System.out.print( "->" );
         node = node.next;
     }
}
  
//Function to get the size of linked list
static int getSize(Node head)
{
     if (head == null )
         return 0 ;
     return 1 + getSize(head.next);
}
  
//Utility function to print the Linked List
//from middle to left right order
static Node printMiddleToLeftRightUtil(Node head, int counter, int lSize)
{
     //Base Condition
     //When size of list is odd
     if (counter == 1 && lSize % 2 != 0 )
     {
  
         //Print node value
         System.out.print( head.data);
  
         //Returns address of next node
         return head.next;
     }
  
     //Base Condition
     //When size of list is even
     else if (counter == 1 )
     {
  
         //Print node value
         //and next node value
         System.out.print(head.data);
         System.out.print( " , " + head.next.data);
  
         //Returns address of next to next node
         return head.next.next;
     }
     else 
     {
  
         //Recursive function call and
         //store return address
         Node ptr = printMiddleToLeftRightUtil(head.next, counter - 1 , lSize);
  
         //Print head data
         System.out.print( " , " + head.data);
  
         //Print ptr data
         System.out.print( " , " + ptr.data);
  
         //Returns address of next node
         return ptr.next;
     }
}
  
//Function to print Middle to
//Left-right order
static void printMiddleToLeftRight(Node head)
{
     //Function call to get the size
     //Of Linked List
     int listSize = getSize(head);
  
     int middle = 0 ;
  
     //Store middle when Linked
     //List size is odd
     if (listSize % 2 != 0 ) 
     {
         middle = (listSize + 1 ) /2 ;
     }
  
     //Store middle when Linked
     //List size is even
     else 
     {
         middle = listSize /2 ;
     }
  
     //Utility function call print
     //Linked List from Middle
     //to left right order
     System.out.print( "Output : " );
  
     printMiddleToLeftRightUtil(head, middle, listSize);
}
  
//Driver code
public static void main(String args[])
{
     //Start with the empty list
     Node head = null ;
  
     //Insert 6. So linked list
     //becomes 6.null
     head = append(head, 6 );
  
     //Insert 6. So linked list
     //becomes 6.4.null
     head = append(head, 4 );
     head = append(head, 8 );
     head = append(head, 7 );
     head = append(head, 9 );
     head = append(head, 11 );
     head = append(head, 2 );
  
     //After inserting linked list
     //becomes 6.4.8.7.9.11.2.null
     System.out.print( "Created Linked list is: " );
  
     //Function to display Linked List content
     printList(head);
     System.out.println();
  
     //Function call print Linked List from
     //Middle to left right order
     printMiddleToLeftRight(head);
}
}
  
//This code is contributed by Arnab Kundu

C#

//A C# program to demonstrate
//the printing of Linked List middle
//to left right order
using System;
  
public class GFG
{
   
//A linked list node
public class Node
{
     public int data;
     public Node next;
};
   
//Given a reference (pointer to pointer)
//to the head of a list and an int, appends
//a new node at the end
static Node append(Node head_ref, int new_data)
{
     //Allocate node
     Node new_node = new Node();
   
     //Used in step 5
     Node last = head_ref;
   
     //Put in the data
     new_node.data = new_data;
   
     //This new node is going to be
     //the last node, so make next of
     //it as null
     new_node.next = null ;
   
     //If the Linked List is empty, //then make the new node as head
     if (head_ref == null )
     {
         head_ref = new_node;
         return head_ref;
     }
   
     //Else traverse till the last node
     while (last.next != null )
         last = last.next;
   
     //Change the next of last node
     last.next = new_node;
     return head_ref;
}
   
//This function prints contents of
//linked list starting from head
static void printList(Node node)
{
     while (node != null ) 
     {
         Console.Write( " " + node.data);
   
         if (node.next != null )
             Console.Write( "->" );
         node = node.next;
     }
}
   
//Function to get the size of linked list
static int getSize(Node head)
{
     if (head == null )
         return 0;
     return 1 + getSize(head.next);
}
   
//Utility function to print the Linked List
//from middle to left right order
static Node printMiddleToLeftRightUtil(Node head, int counter, int lSize)
{
     //Base Condition
     //When size of list is odd
     if (counter == 1 && lSize % 2 != 0)
     {
   
         //Print node value
         Console.Write( head.data);
   
         //Returns address of next node
         return head.next;
     }
   
     //Base Condition
     //When size of list is even
     else if (counter == 1)
     {
   
         //Print node value
         //and next node value
         Console.Write(head.data);
         Console.Write( " , " + head.next.data);
   
         //Returns address of next to next node
         return head.next.next;
     }
     else
     {
   
         //Recursive function call and
         //store return address
         Node ptr = printMiddleToLeftRightUtil(head.next, counter - 1, lSize);
   
         //Print head data
         Console.Write( " , " + head.data);
   
         //Print ptr data
         Console.Write( " , " + ptr.data);
   
         //Returns address of next node
         return ptr.next;
     }
}
   
//Function to print Middle to
//Left-right order
static void printMiddleToLeftRight(Node head)
{
     //Function call to get the size
     //Of Linked List
     int listSize = getSize(head);
   
     int middle = 0;
   
     //Store middle when Linked
     //List size is odd
     if (listSize % 2 != 0) 
     {
         middle = (listSize + 1) /2;
     }
   
     //Store middle when Linked
     //List size is even
     else
     {
         middle = listSize /2;
     }
   
     //Utility function call print
     //Linked List from Middle
     //to left right order
     Console.Write( "Output : " );
   
     printMiddleToLeftRightUtil(head, middle, listSize);
}
   
//Driver code
public static void Main(String []args)
{
     //Start with the empty list
     Node head = null ;
   
     //Insert 6. So linked list
     //becomes 6.null
     head = append(head, 6);
   
     //Insert 6. So linked list
     //becomes 6.4.null
     head = append(head, 4);
     head = append(head, 8);
     head = append(head, 7);
     head = append(head, 9);
     head = append(head, 11);
     head = append(head, 2);
   
     //After inserting linked list
     //becomes 6.4.8.7.9.11.2.null
     Console.Write( "Created Linked list is: " );
   
     //Function to display Linked List content
     printList(head);
     Console.WriteLine();
   
     //Function call print Linked List from
     //Middle to left right order
     printMiddleToLeftRight(head);
}
}
  
//This code is contributed by Rajput-Ji

输出如下:

Created Linked list is:  6-> 4-> 8-> 7-> 9-> 11-> 2
Output : 7 , 8 , 9 , 4 , 11 , 6 , 2

木子山

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