在链表中找到最大和第二大的值

2021年5月16日17:10:45 发表评论 1,080 次浏览

本文概述

给定一个链表, 任务是在链表中找到最大和第二大的值。

例子：

输入：LL = 10-> 15-> 5-> 20-> 7-> 9
输出：最大= 20
第二最大= 15
输入：LL = 0-> 5-> 52-> 21
输出：最大= 52
第二最大= 21

方法：

  1. 将前两个节点的最大值存储在变量中最高.
  2. 将前两个节点中的最小值存储在变量中second_max.
  3. 遍历其余的链表。对于每个节点：
    • 如果当前节点的值大于max, 则将second_max设置为max, 将max设置为当前节点的值。
    • 否则, 如果当前节点的值大于second_max, 则将second_max设置为当前节点的值。

下面是上述方法的实现：

C ++

//C++ program to find the largest and
//second largest element in a Linked List
  
#include <bits/stdc++.h>
using namespace std;
  
//Link list node
struct Node {
     int data;
     struct Node* next;
};
  
//Function to push the node at the
//beginning of the linked list
void push( struct Node** head_ref, int new_data)
{
     struct Node* new_node
         = ( struct Node*) malloc (
             sizeof ( struct Node));
  
     new_node->data = new_data;
     new_node->next = (*head_ref);
     (*head_ref) = new_node;
}
  
//Function to print the largest
//and second largest element
void findLargestAndSecondLargest( struct Node* head)
{
     //initialise max and second max using
     //first two nodes of linked list
     int val1 = head->data, val2 = head->next->data, max = std::max(val1, val2), second_max = std::min(val1, val2);
  
     //move the head pointer to 3rd node
     head = head->next->next;
  
     //iterate over rest of linked list
     while (head != NULL) {
  
         if (head->data> max) {
  
             //If current node value is greater
             //than max, then set second_max as
             //current max value and max as
             //current node value
             second_max = max;
             max = head->data;
         }
         else if (head->data> second_max) {
  
             //else if current node value is
             //greater than second_max, set
             //second_max as node value
             second_max = head->data;
         }
  
         //move the head pointer to next node
         head = head->next;
     }
  
     //Print the largest
     //and second largest value
     cout <<"Largest = "
          <<max <<endl;
     cout <<"Second Largest = "
          <<second_max <<endl;
}
  
//Driver code
int main()
{
     struct Node* head = NULL;
  
     push(&head, 20);
     push(&head, 5);
     push(&head, 15);
     push(&head, 10);
     push(&head, 7);
     push(&head, 6);
     push(&head, 11);
     push(&head, 9);
  
     findLargestAndSecondLargest(head);
  
     return 0;
}

Java

//Java program to find the largest and
//second largest element in a Linked List
class GFG{
   
//Link list node
static class Node {
     int data;
     Node next;
  
};
   
//Function to push the node at the
//beginning of the linked list
static Node push(Node head_ref, int new_data)
{
     Node new_node
         = new Node();
   
     new_node.data = new_data;
     new_node.next = head_ref;
     head_ref = new_node;
     return head_ref;
}
   
//Function to print the largest
//and second largest element
static void findLargestAndSecondLargest(Node head)
{
     //initialise max and second max using
     //first two nodes of linked list
     int val1 = head.data, val2 = head.next.data, max = Math.max(val1, val2), second_max = Math.min(val1, val2);
   
     //move the head pointer to 3rd node
     head = head.next.next;
   
     //iterate over rest of linked list
     while (head != null ) {
   
         if (head.data> max) {
   
             //If current node value is greater
             //than max, then set second_max as
             //current max value and max as
             //current node value
             second_max = max;
             max = head.data;
         }
         else if (head.data> second_max) {
   
             //else if current node value is
             //greater than second_max, set
             //second_max as node value
             second_max = head.data;
         }
   
         //move the head pointer to next node
         head = head.next;
     }
   
     //Print the largest
     //and second largest value
     System.out.print( "Largest = "
          + max + "\n" );
     System.out.print( "Second Largest = "
          + second_max + "\n" );
}
   
//Driver code
public static void main(String[] args)
{
     Node head = null ;
   
     head = push(head, 20 );
     head = push(head, 5 );
     head = push(head, 15 );
     head = push(head, 10 );
     head = push(head, 7 );
     head = push(head, 6 );
     head = push(head, 11 );
     head = push(head, 9 );
   
     findLargestAndSecondLargest(head);
}
}
  
//This code is contributed by Rajput-Ji

Python3

# Python3 program to find the largest and
# second largest element in a Linked List
# Node class 
class Node: 
      
     # Function to initialize the node object 
     def __init__( self , data): 
          
         # Assign data 
         self .data = data 
          
         # Initialize 
         # next as null
         self . next = None 
      
# Linked List Class
class LinkedList:
  
     # Function to initialize the
     # LinkedList class.
     def __init__( self ): 
          
         # Initialize head as None 
         self .head = None 
  
     # This function insert a new node at the 
     # beginning of the linked list 
     def push( self , new_data): 
      
         # Create a new Node 
         new_node = Node(new_data) 
  
         # Make next of new Node as head 
         new_node. next = self .head 
  
         # Move the head to point to new Node 
         self .head = new_node
  
     # Function to find the max and
     # second largest value from the list
     def findLargestAndSecondLargest( self ):
  
         # Take a Head to iterate list
         Head = self .head
  
         # Initialize max and second_max
         # using first two nodes of the list
         val1 = Head.data
         val2 = Head. next .data
         Max = max (val1, val2)
         second_max = min (val1, val2)
  
         # Move the Head to third node
         Head = Head. next . next
  
         # Iterate over rest of linked list
         while (Head ! = None ):
  
             # If current node value is 
             # greater then Max then
             if (Head.data> Max ):
  
                 # Set the current max to second_max
                 # and current node value to max
                 second_max = Max
                 Max = Head.data
  
             # Else if current node value is
             # greater then second_max value
             elif (Head.data> second_max):
  
                 # Then current node value
                 # to second_max
                 second_max = Head.data
  
             # Move the head to next node
             Head = Head. next
  
         # Print the largest and second largest values
         print ( "Largest = " , Max )
         print ( "Second Largest = " , second_max)
  
# Driver code
if __name__ = = '__main__' :
  
     # Initialising the linked list
     head = LinkedList()
  
     # Pushing the values in list
     head.push( 20 )
     head.push( 5 )
     head.push( 15 )
     head.push( 10 )
     head.push( 7 )
     head.push( 6 )
     head.push( 11 )
     head.push( 9 )
      
     # Calling the function to print
     # largest and second largest values.
     head.findLargestAndSecondLargest()
              
# This code is contributed by Amit Mangal.

C#

//C# program to find the largest and
//second largest element in a Linked List
using System;
  
class GFG{
    
//Link list node
class Node {
     public int data;
     public Node next;
   
};
    
//Function to push the node at the
//beginning of the linked list
static Node push(Node head_ref, int new_data)
{
     Node new_node
         = new Node();
    
     new_node.data = new_data;
     new_node.next = head_ref;
     head_ref = new_node;
     return head_ref;
}
    
//Function to print the largest
//and second largest element
static void findLargestAndSecondLargest(Node head)
{
     //initialise max and second max using
     //first two nodes of linked list
     int val1 = head.data, val2 = head.next.data, max = Math.Max(val1, val2), second_max = Math.Min(val1, val2);
    
     //move the head pointer to 3rd node
     head = head.next.next;
    
     //iterate over rest of linked list
     while (head != null ) {
    
         if (head.data> max) {
    
             //If current node value is greater
             //than max, then set second_max as
             //current max value and max as
             //current node value
             second_max = max;
             max = head.data;
         }
         else if (head.data> second_max) {
    
             //else if current node value is
             //greater than second_max, set
             //second_max as node value
             second_max = head.data;
         }
    
         //move the head pointer to next node
         head = head.next;
     }
    
     //Print the largest
     //and second largest value
     Console.Write( "Largest = "
          + max + "\n" );
     Console.Write( "Second Largest = "
          + second_max + "\n" );
}
    
//Driver code
public static void Main(String[] args)
{
     Node head = null ;
    
     head = push(head, 20);
     head = push(head, 5);
     head = push(head, 15);
     head = push(head, 10);
     head = push(head, 7);
     head = push(head, 6);
     head = push(head, 11);
     head = push(head, 9);
    
     findLargestAndSecondLargest(head);
}
}
  
//This code is contributed by Princi Singh

输出如下：

Largest = 20
Second Largest = 15

绩效分析:

  • 时间复杂度：在上述方法中, 由于我们仅对链表进行迭代一次, 因此时间复杂度为上).
  • 辅助空间复杂度：在上述方法中, 除了几个恒定大小的变量之外, 我们没有使用任何额外的空间, 因此辅助空间的复杂度为O(1).
木子山

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