C 中 n 数组的交集和并集

## 数组交集与并集的生成

c
#include 
void intersection(int arr1[], int arr2[], int m, int n) {
    printf("数组交集：");
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (arr1[i] == arr2[j]) {
                printf("%d ", arr1[i]);
                break;
            }
        }
    }
}
int main() {
    int arr1[] = {1, 2, 3, 4, 5};
    int arr2[] = {3, 4, 5, 6, 7};
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    intersection(arr1, arr2, m, n);
    return 0;
}

c
#include 
void unionSet(int arr1[], int arr2[], int m, int n) {
    printf("%
数组并集：");
    int mergedArr[m + n];
    int mergedIndex = 0;
    for (int i = 0; i < m; i++) {
        mergedArr[mergedIndex++] = arr1[i];
    }
    for (int i = 0; i < n; i++) {
        int flag = 0;
        for (int j = 0; j < m; j++) {
            if (arr2[i] == arr1[j]) {
                flag = 1;
                break;
            }
        }
        if (!flag) {
            mergedArr[mergedIndex++] = arr2[i];
        }
    }
    for (int i = 0; i < mergedIndex; i++) {
        printf("%d ", mergedArr[i]);
    }
}
int main() {
    int arr1[] = {1, 2, 3, 4, 5};
    int arr2[] = {3, 4, 5, 6, 7};
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    unionSet(arr1, arr2, m, n);
    return 0;
}