c++求两个数组的并集

可以使用STL中的set来求两个数组的并集，具体代码如下：

#include <iostream>
#include <set>
using namespace std;

int main() {
    int arr1[] = {1, 2, 3, 4, 5};
    int arr2[] = {3, 4, 5, 6, 7};
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr2) / sizeof(arr2[0]);

    set<int> s;
    for(int i = 0; i < n1; i++) {
        s.insert(arr1[i]);
    }
    for(int i = 0; i < n2; i++) {
        s.insert(arr2[i]);
    }

    cout << "两个数组的并集为：";
    for(auto it = s.begin(); it != s.end(); it++) {
        cout << *it << " ";
    }
    cout << endl;

    return 0;
}

输出结果为：

两个数组的并集为：1 2 3 4 5 6 7 

相关问题

c++求两个数组的并集和交集，能实现多组数据的多次运算，直到选择结束运算为止，提供问答界面，以便人机交互

下面是一个示例代码，实现了多组数据的多次运算，直到选择结束运算为止，并提供了问答界面，以便人机交互：

#include <iostream>
#include <algorithm>
#include <set>
#include <vector>

using namespace std;

void printArray(int arr[], int n)
{
    for (int i = 0; i < n; ++i) {
        cout << arr[i] << " ";
    }
    cout << endl;
}

void printSet(set<int> s)
{
    for (auto it = s.begin(); it != s.end(); ++it) {
        cout << *it << " ";
    }
    cout << endl;
}

void printIntersectionUnion(int arr1[], int n1, int arr2[], int n2)
{
    set<int> set1(arr1, arr1 + n1);
    set<int> set2(arr2, arr2 + n2);

    vector<int> intersection(n1 + n2);
    auto it = set_intersection(set1.begin(), set1.end(), set2.begin(), set2.end(), intersection.begin());
    intersection.resize(it - intersection.begin());

    vector<int> uni(n1 + n2);
    it = set_union(set1.begin(), set1.end(), set2.begin(), set2.end(), uni.begin());
    uni.resize(it - uni.begin());

    cout << "Intersection: ";
    printArray(intersection.data(), intersection.size());
    cout << "Union: ";
    printSet(set<int>(uni.begin(), uni.end()));
}

int main()
{
    int arr1[100], arr2[100], n1, n2;
    char choice;

    do {
        cout << "Please enter the size of array 1: ";
        cin >> n1;
        cout << "Please enter the elements of array 1: ";
        for (int i = 0; i < n1; ++i) {
            cin >> arr1[i];
        }

        cout << "Please enter the size of array 2: ";
        cin >> n2;
        cout << "Please enter the elements of array 2: ";
        for (int i = 0; i < n2; ++i) {
            cin >> arr2[i];
        }

        printIntersectionUnion(arr1, n1, arr2, n2);

        cout << "Do you want to continue? (y/n): ";
        cin >> choice;
    } while (choice == 'y' || choice == 'Y');

    cout << "Program ended." << endl;

    return 0;
}

示例输出：

Please enter the size of array 1: 5
Please enter the elements of array 1: 1 2 3 4 5
Please enter the size of array 2: 5
Please enter the elements of array 2: 3 4 5 6 7
Intersection: 3 4 5 
Union: 1 2 3 4 5 6 7 
Do you want to continue? (y/n): y
Please enter the size of array 1: 3
Please enter the elements of array 1: 1 2 3
Please enter the size of array 2: 4
Please enter the elements of array 2: 2 3 4 5
Intersection: 2 3 
Union: 1 2 3 4 5 
Do you want to continue? (y/n): n
Program ended.

c++数组求并集

假设有两个数组A和B，它们的长度分别为n和m，可以使用以下代码求它们的并集：

#include <iostream>
using namespace std;

const int MAXN = 100;

int main() {
    int n, m;
    int A[MAXN], B[MAXN], C[MAXN];  // 定义三个数组
    cin >> n;                       // 输入数组A的长度
    for (int i = 0; i < n; i++) {
        cin >> A[i];                // 输入数组A的元素
    }
    cin >> m;                       // 输入数组B的长度
    for (int i = 0; i < m; i++) {
        cin >> B[i];                // 输入数组B的元素
    }
    int k = 0;                      // 计数器
    for (int i = 0; i < n; i++) {   // 遍历数组A
        bool flag = true;           // 定义一个标志
        for (int j = 0; j < m; j++) {   // 遍历数组B
            if (A[i] == B[j]) {     // 如果A[i]等于B[j]
                flag = false;       // 标志置为false
                break;              // 跳出循环
            }
        }
        if (flag) {                 // 如果标志为true
            C[k++] = A[i];          // 将A[i]添加到C中，并将计数器加1
        }
    }
    for (int i = 0; i < m; i++) {   // 遍历数组B
        bool flag = true;           // 定义一个标志
        for (int j = 0; j < n; j++) {   // 遍历数组A
            if (B[i] == A[j]) {     // 如果B[i]等于A[j]
                flag = false;       // 标志置为false
                break;              // 跳出循环
            }
        }
        if (flag) {                 // 如果标志为true
            C[k++] = B[i];          // 将B[i]添加到C中，并将计数器加1
        }
    }
    cout << "A和B的并集为：";
    for (int i = 0; i < k; i++) {   // 遍历数组C
        cout << C[i] << " ";        // 输出数组C的元素
    }
    cout << endl;
    return 0;
}

该程序的思路是：遍历数组A，对于每个元素，如果它在数组B中出现过，则跳过；否则将它添加到数组C中。然后遍历数组B，对于每个元素，如果它在数组A中出现过，则跳过；否则将它添加到数组C中。最后输出数组C即可得到两个数组的并集。