C++用集合C返回集合A与集合B的差集

集合C返回集合A与集合B的差集可以使用以下代码实现：

#include <stdio.h>

// 定义集合的最大大小
#define MAX_SIZE 100

// 函数声明
int* difference(int arrA[], int sizeA, int arrB[], int sizeB, int* sizeC);

int main() {
    int setA[] = {1, 2, 3, 4, 5};
    int sizeA = sizeof(setA) / sizeof(int);
    
    int setB[] = {3, 4, 5, 6};
    int sizeB = sizeof(setB) / sizeof(int);
    
    int setC[MAX_SIZE];
    int sizeC;
    
    int* result = difference(setA, sizeA, setB, sizeB, &sizeC);
    
    printf("集合A与集合B的差集为: ");
    for (int i = 0; i < sizeC; i++) {
        printf("%d ", result[i]);
    }
    printf("\n");
    
    return 0;
}

int* difference(int arrA[], int sizeA, int arrB[], int sizeB, int* sizeC) {
    int setC[MAX_SIZE];
    int index = 0;
    
    for (int i = 0; i < sizeA; i++) {
        int found = 0;
        
        for (int j = 0; j < sizeB; j++) {
            if (arrA[i] == arrB[j]) {
                found = 1;
                break;
            }
        }
        
        if (!found) {
            setC[index++] = arrA[i];
        }
    }
    
    *sizeC = index;
    return setC;
}