./Solution.cpp: In member function 'std::vector<int> Solution::findUnion(int*, int*, int, int)':./Solution.cpp:16:19: error: 'begin' was not declared in this scope for(int i:arr1){ ^./Solution.cpp:16:19: note: suggested alternatives:In file included from /usr/local/include/c++/5.4.0/x86_64-linux-gnu/bits/stdc++.h:94:0, from ./Solution.cpp:2:/usr/local/include/c++/5.4.0/valarray:1206:5: note: 'std::begin' begin(const valarray<_Tp>& __va)
Question
./Solution.cpp: In member function 'std::vector<int> Solution::findUnion(int*, int*, int, int)':./Solution.cpp:16:19: error: 'begin' was not declared in this scope for(int i:arr1){ ^./Solution.cpp:16:19: note: suggested alternatives:In file included from /usr/local/include/c++/5.4.0/x86_64-linux-gnu/bits/stdc++.h:94:0, from ./Solution.cpp:2:/usr/local/include/c++/5.4.0/valarray:1206:5: note: 'std::begin' begin(const valarray<_Tp>& __va)
Solution
El error que estás viendo se debe a que arr1 no es un contenedor que soporte el rango basado en bucles for en la forma en que lo estás utilizando. Aquí hay una forma de corregir el código paso a paso:
- Asegúrate de que
arr1yarr2sean vectores o cualquier otro contenedor que soporte iteradores. - Si
arr1yarr2son arreglos, necesitarás convertirlos a vectores antes de usarlos en un bucleforbasado en rango.
Aquí hay un ejemplo de cómo podrías modificar tu código:
#include <vector>
#include <algorithm>
#include <set>
class Solution {
public:
std::vector<int> findUnion(int* arr1, int* arr2, int n, int m) {
std::set<int> unionSet;
// Insert elements of arr1 into the set
for(int i = 0; i < n; ++i) {
unionSet.insert(arr1[i]);
}
// Insert elements of arr2 into the set
for(int i = 0; i < m; ++i) {
unionSet.insert(arr2[i]);
}
// Convert set to vector
std::vector<int> unionVector(unionSet.begin(), unionSet.end());
return unionVector;
}
};
En este ejemplo, arr1 y arr2 son arreglos de enteros, y n y m son sus respectivos tamaños. Utilizamos un std::set para almacenar los elementos únicos de ambos arreglos y luego convertimos el set a un vector para devolver el resultado.
Este enfoque asegura que no haya duplicados en la unión y que los elementos estén ordenados.
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