programa en C ++ para encontrar Unión de dos listas enlazadas simples

En este artículo, vamos a ver cómo encontrar unión de dos lista enlazada de manera eficiente el uso de combinar la clasificación ?

Planteamiento del problema: Escribir un programa en C ++ para encontrar la unión de dos listas enlazadas simples.

Ejemplo:

    Let the first linked list be:
5->4->3->6->1->NULL
Let the second linked list to be:
3->8->9->12->1->NULL
So there union is:
5->4->3->6->1->8->9->12->NULL

enfoque Solución

fuerza bruta:

Una técnica puede ser a class todos los nodos de una lista enlazada y include otra comprobación vinculado si los nodos ya están adjunta o no. Este enfoque lleva O (m * n) veces complejidad . m , n = longitud de listas enlazadas.

enfoque eficiente:

utilización enfoque eficiente para utilizar de combinación de tipo .

  1. Ordenar tanto la lista enlazada mediante fusión especie. (for detallada, consulte: Combinar ordenar for individuales listas enlazadas)
  2. Scan cada lista y acumulación de unión ligado de acuerdo con lo siguiente:
  3. IF (list1->data < list2->data)
    Createnode(list1->data) && append node to the union list
    Traverse list1 by one step( list1=list1->next)
    ELSE IF(list1->data ==list2->data)
    Createnode(list1->data) && append nodeto the union list
    Traverse list1 by one step( list1=list1->next)
    Traverse list2 by one step( list2=list2->next)
    ELSE//list1->data >list2->data
    Createnode(list2->data) && append node to the union list
    Traverse list2 by one step( list2=list2->next)
    END IF-ELSE

  4. Mostrar la lista de la Unión

implementación en C ++ para encontrar Unión de dos listas enlazadas individuales

#include<bits/stdc++.h>
using namespace std;
class node{
    public:
int data; // data field
struct node *next;
};
void display(class node* head){
node* current=head; // current node set to head
while(current!=NULL){ //traverse until current node isn't NULL
printf("%d ",current->data);
current=current->next; // go to next node
}
}
node* creatnode(int d){
node* temp=(node*)malloc(sizeof(node));
temp->data=d;
temp->next=NULL;
return temp;
}
node* mergeList(node* split1,node* split2){
node* newhead=NULL;
//base cases
if(split1==NULL)
return split2;
if(split2==NULL)
return split1;
if(split1->data<=split2->data){
newhead=split1;
newhead->next=mergeList(split1->next,split2);
}
else{
newhead=split2;
newhead->next=mergeList(split1,split2->next);
}
return newhead;
}
void splitList(node* head,node** split1,node** split2){
node* slow=head;
node* fast=head->next;
while(fast!=NULL){
fast=fast->next;
if(fast!=NULL){
slow=slow->next;
fast=fast->next;
}
}
*split1=head;
*split2=slow->next;
//spliting
slow->next=NULL;
}
void mergeSort(node** refToHead){
node* head=*refToHead;
node* split1,*split2;
//base case
if(head==NULL || head->next==NULL){
return;
}
//split the list in two halves
splitList(head,&split1,&split2);
//recursively sort the two halves
mergeSort(&split1);
mergeSort(&split2);
*refToHead=mergeList(split1,split2);
return;
}
node* findUnion(node* head1, node* head2){
if(head1==NULL && head2==NULL)
return NULL;
node* head3;
if(head1->data<head2->data){
head3=creatnode(head1->data);
head1=head1->next;
}
else if(head1->data==head2->data){
head3=creatnode(head1->data);
head1=head1->next;
head2=head2->next;
}
else{
head3=creatnode(head2->data);
head2=head2->next;
}
node* temp=head3;
while( head1!=NULL && head2!=NULL){
if(head1->data<head2->data){
temp->next=creatnode(head1->data);
temp=temp->next;
head1=head1->next;
}
else if(head1->data==head2->data){
temp->next=creatnode(head1->data);
temp=temp->next;
head1=head1->next;
head2=head2->next;
}
else{
temp->next=creatnode(head2->data);
temp=temp->next;
head2=head2->next;
}
}
while(head1!=NULL){
temp->next=creatnode(head1->data);
temp=temp->next;
head1=head1->next;
}
while(head2!=NULL){
temp->next=creatnode(head2->data);
temp=temp->next;
head2=head2->next;
}
return head3;
}
int main(){
printf("creating the linked list by inserting new nodes at the endn");
printf("enter 0 to stop building the list, else enter any integern");
int k;
node* curr,*temp;
cout<<"enter list1...n";
scanf("%d",&k);
node* head1=creatnode(k); //buliding list, first node
scanf("%d",&k);
temp=head1;
///////////////////inserting at the end//////////////////////
while(k){
curr=creatnode(k);
temp->next=curr;//appending each node
temp=temp->next;
scanf("%d",&k);
}
cout<<"displaying list1...n";
display(head1); // displaying the list
cout<<"nenter list2...n";
scanf("%d",&k);
node* head2=creatnode(k); //buliding list, first node
scanf("%d",&k);
temp=head2;
///////////////////inserting at the end//////////////////////
while(k){
curr=creatnode(k);
temp->next=curr;//appending each node
temp=temp->next;
scanf("%d",&k);
}
cout<<"displaying list1...n";
display(head2);
//sort both the lists
mergeSort(&head1);
mergeSort(&head2);
cout<<"ndisplaying their union...n";
node* head3=findUnion(head1,head2);
display(head3);
return 0;
}

salida

creating the linked list by inserting new nodes at the end
enter 0 to stop building the list, else enter any integer
enter list1...
5 4 3 6 1 0
displaying list1...
5 4 3 6 1
enter list2...
3 8 9 12 1 0
displaying list1...
3 8 9 12 1
displaying their union...
1 3 4 5 6 8 9 12

Ejemplo con la explicación:

First linked list: 5->4->3->6->1->NULL
Second linked list:3->8->9->12->1->NULL
After applying merge sort:
First linked list: 1->3->4->5->6->NULL
Second linked list: 1->3->8->9->12->NULL
---------------------------------------------------------
//for better understanding nodes are represented only by respective values
head1=1
head2=1
FindUnion(1,1):
0th iteration
head1!=NULL&&head2!=NULL
head1->data==head2->data //=1
thus head3(head of union list) =1
temp=head3=1
union list up to now:
1->NULL
head1=1->next=3;
head2=1->next=3;
1st iteration
head1!=NULL&&head2!=NULL
head1->data==head2->data //=3
thus temp->next =3
temp=temp->next=3;
union list up to now:
1->3->NULL
head1=3->next=4;
head2=3->next=8;
2nd iteration
head1!=NULL&&head2!=NULL
head1->data<head2->data //4<8
thus temp->next =head1->data=4
temp=temp->next=4;
union list up to now:
1->3->4->NULL
head1=4->next=5;
3rd iteration
head1!=NULL&&head2!=NULL
head1->data<head2->data //5<8
thus temp->next =head1->data=5
temp=temp->next=5;
union list up to now:
1->3->4->5->NULL
head1=5->next=6;
4th iteration
head1!=NULL&&head2!=NULL
head1->data<head2->data //6<8
thus temp->next =head1->data=6
temp=temp->next=6;
union list up to now:
1->3->4->5->6->NULL
head1=6->next=NULL;
5th iteration
head1 ==NULL
head2!=NULL
thus temp->next =head2->data=8
temp=temp->next=8;
union list up to now:
1->3->4->5->6->8->NULL
head2=8->next=9;
6th iteration
head1 ==NULL
head2!=NULL
thus temp->next =head2->data=9
temp=temp->next=9;
union list up to now:
1->3->4->5->6->8->9->NULL
head2=9->next=12;
7th iteration
head1 ==NULL
head2!=NULL
thus temp->next =head2->data=12
temp=temp->next=12;
union list up to now: 1->3->4->5->6->8->9->12->NULL
head2=12->next=NULL;
8th iteration:
head1==NULL
head2==NULL
thus no more scanning, union list is built
union: 1->3->4->5->6->8->9->12->NULL
Head3 at 1


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