Dados dos arreglos, encuentre elementos comunes entre estos dos arreglos usando STL en C++.
Ejemplo:
Input:
arr1[] = {1, 45, 54, 71, 76, 12},
arr2[] = {1, 7, 5, 4, 6, 12}
Output: {1, 12}
Input:
arr1[] = {1, 7, 5, 4, 6, 12},
arr2[] = {10, 12, 11}
Output: {1, 4, 12}
Enfoque: los elementos comunes se pueden encontrar con la ayuda de la función set_intersection() proporcionada en STL.
Sintaxis:
set_intersection (InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2, InputIterator2 last2,
OutputIterator result);
// C++ program to find common elements
// between two Arrays using STL
#include <bits/stdc++.h>
using namespace std;
int main()
{
// Get the array
int arr1[] = { 1, 45, 54, 71, 76, 12 };
int arr2[] = { 1, 7, 5, 4, 6, 12 };
// Compute the sizes
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int n2 = sizeof(arr2) / sizeof(arr2[0]);
// Sort the arrays
sort(arr1, arr1 + n1);
sort(arr2, arr2 + n2);
// Print the array
cout << "First Array: ";
for (int i = 0; i < n1; i++)
cout << arr1[i] << " ";
cout << endl;
cout << "Second Array: ";
for (int i = 0; i < n2; i++)
cout << arr2[i] << " ";
cout << endl;
// Initialise a vector
// to store the common values
// and an iterator
// to traverse this vector
vector<int> v(n1 + n2);
vector<int>::iterator it, st;
it = set_intersection(arr1, arr1 + n1,
arr2, arr2 + n2,
v.begin());
cout << "\nCommon elements:\n";
for (st = v.begin(); st != it; ++st)
cout << *st << ", ";
cout << '\n';
return 0;
}
Producción:
First Array: 1 12 45 54 71 76 Second Array: 1 4 5 6 7 12 Common elements: 1, 12,