Dada una lista enlazada de datos enteros. La tarea es escribir un programa que encuentre de manera eficiente el segundo elemento más grande presente en la Lista enlazada.
Ejemplos :
Input : List = 12 -> 35 -> 1 -> 10 -> 34 -> 1 Output : The second largest element is 34. Input : List = 10 -> 5 -> 10 Output : The second largest element is 5.
Una solución simple será ordenar primero la lista vinculada en orden descendente y luego imprimir el segundo elemento de la lista vinculada ordenada. La complejidad temporal de esta solución es O(nlogn).
Una solución mejor es recorrer la lista Vinculada dos veces. En el primer recorrido encuentre el elemento máximo. En el segundo recorrido encuentre el elemento mayor menor que el elemento obtenido en el primer recorrido. La complejidad temporal de esta solución es O(n).
Una solución más eficiente puede ser encontrar el segundo elemento más grande en un solo recorrido.
A continuación se muestra el algoritmo completo para hacer esto:
1) Initialize two variables first and second to INT_MIN as,
first = second = INT_MIN
2) Start traversing the Linked List,
a) If the current element in Linked List say list[i] is greater
than first. Then update first and second as,
second = first
first = list[i]
b) If the current element is in between first and second,
then update second to store the value of current variable as
second = list[i]
3) Return the value stored in second node.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to print second largest
// value in a linked list
#include <climits>
#include <iostream>
using namespace std;
// A linked list node
struct Node {
int data;
struct Node* next;
};
// Function to add a node at the
// beginning of Linked List
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node = (struct Node*)malloc(sizeof(struct 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;
}
// Function to count size of list
int listSize(struct Node* node)
{
int count = 0;
while (node != NULL) {
count++;
node = node->next;
}
return count;
}
// Function to print the second
// largest element
void print2largest(struct Node* head)
{
int i, first, second;
int list_size = listSize(head);
/* There should be atleast two elements */
if (list_size < 2) {
cout << "Invalid Input";
return;
}
first = second = INT_MIN;
struct Node* temp = head;
while (temp != NULL) {
if (temp->data > first) {
second = first;
first = temp->data;
}
// If current node's data is in between
// first and second then update second
else if (temp->data > second && temp->data != first)
second = temp->data;
temp = temp->next;
}
if (second == INT_MIN)
cout << "There is no second largest element\n";
else
cout << "The second largest element is " << second;
}
// Driver program to test above function
int main()
{
struct Node* start = NULL;
/* The constructed linked list is:
12 -> 35 -> 1 -> 10 -> 34 -> 1 */
push(&start, 1);
push(&start, 34);
push(&start, 10);
push(&start, 1);
push(&start, 35);
push(&start, 12);
print2largest(start);
return 0;
}
Java
// Java program to print second largest
// value in a linked list
class GFG
{
// A linked list node
static class Node
{
int data;
Node next;
};
// Function to add a node at the
// beginning of 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 count size of list
static int listSize( Node node)
{
int count = 0;
while (node != null)
{
count++;
node = node.next;
}
return count;
}
// Function to print the second
// largest element
static void print2largest( Node head)
{
int i, first, second;
int list_size = listSize(head);
// There should be atleast two elements /
if (list_size < 2)
{
System.out.print("Invalid Input");
return;
}
first = second = Integer.MIN_VALUE;
Node temp = head;
while (temp != null)
{
if (temp.data > first)
{
second = first;
first = temp.data;
}
// If current node's data is in between
// first and second then update second
else if (temp.data > second && temp.data != first)
second = temp.data;
temp = temp.next;
}
if (second == Integer.MIN_VALUE)
System.out.print( "There is no second largest element\n");
else
System.out.print ("The second largest element is " + second);
}
// Driver program to test above function
public static void main(String args[])
{
Node start = null;
// The constructed linked list is:
//12 . 35 . 1 . 10 . 34 . 1
start=push(start, 1);
start=push(start, 34);
start=push(start, 10);
start=push(start, 1);
start=push(start, 35);
start=push(start, 12);
print2largest(start);
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 program to print second largest
# value in a linked list
# A linked list node
class Node :
def __init__(self):
self.data = 0
self.next = None
# Function to add a node at the
# beginning of Linked List
def push( head_ref, new_data) :
# allocate node /
new_node = 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 count size of list
def listSize( node):
count = 0
while (node != None):
count = count + 1
node = node.next
return count
# Function to print the second
# largest element
def print2largest( head):
i = 0
first = 0
second = 0
list_size = listSize(head)
# There should be atleast two elements /
if (list_size < 2) :
print("Invalid Input")
return
first = second = -323767
temp = head
while (temp != None):
if (temp.data > first) :
second = first
first = temp.data
# If current node's data is in between
# first and second then update second
elif (temp.data > second and temp.data != first) :
second = temp.data
temp = temp.next
if (second == -323767) :
print( "There is no second largest element\n")
else:
print ("The second largest element is " , second)
# Driver code
start = None
# The constructed linked list is:
# 12 . 35 . 1 . 10 . 34 . 1
start = push(start, 1)
start = push(start, 34)
start = push(start, 10)
start = push(start, 1)
start = push(start, 35)
start = push(start, 12)
print2largest(start)
# This code is contributed by Arnab Kundu
C#
// C# program to print second largest
// value in a linked list
using System;
class GFG
{
// A linked list node
public class Node
{
public int data;
public Node next;
};
// Function to add a node at the
// beginning of 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 count size of list
static int listSize( Node node)
{
int count = 0;
while (node != null)
{
count++;
node = node.next;
}
return count;
}
// Function to print the second
// largest element
static void print2largest(Node head)
{
int first, second;
int list_size = listSize(head);
// There should be atleast two elements
if (list_size < 2)
{
Console.Write("Invalid Input");
return;
}
first = second = int.MinValue;
Node temp = head;
while (temp != null)
{
if (temp.data > first)
{
second = first;
first = temp.data;
}
// If current node's data is in between
// first and second then update second
else if (temp.data > second &&
temp.data != first)
second = temp.data;
temp = temp.next;
}
if (second == int.MinValue)
Console.Write( "There is no second" +
" largest element\n");
else
Console.Write("The second largest " +
"element is " + second);
}
// Driver Code
public static void Main(String []args)
{
Node start = null;
// The constructed linked list is:
//12 . 35 . 1 . 10 . 34 . 1
start = push(start, 1);
start = push(start, 34);
start = push(start, 10);
start = push(start, 1);
start = push(start, 35);
start = push(start, 12);
print2largest(start);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript program to print second largest
// value in a linked list
// A linked list node
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}
// Function to add a node at the
// beginning of Linked List
function push(head_ref, new_data) {
// allocate node
var 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 count size of list
function listSize(node) {
var count = 0;
while (node != null) {
count++;
node = node.next;
}
return count;
}
// Function to print the second
// largest element
function print2largest(head) {
var first, second;
var list_size = listSize(head);
// There should be atleast two elements
if (list_size < 2) {
document.write("Invalid Input");
return;
}
first = second = -2147483648;
var temp = head;
while (temp != null) {
if (temp.data > first) {
second = first;
first = temp.data;
}
// If current node's data is in between
// first and second then update second
else if (temp.data > second && temp.data != first)
second = temp.data;
temp = temp.next;
}
if (second == -2147483648)
document.write("There is no second" + " largest element<br>");
else
document.write("The second largest " + "element is " + second);
}
// Driver Code
var start = null;
// The constructed linked list is:
//12 . 35 . 1 . 10 . 34 . 1
start = push(start, 1);
start = push(start, 34);
start = push(start, 10);
start = push(start, 1);
start = push(start, 35);
start = push(start, 12);
print2largest(start);
</script>
Producción:
The second largest element is 34