C语言实现思路

1. 合并两个给出的数组
2. 将合并后的数组按照从小到大的顺序进行排列
3. 半段n1+n2个数的奇偶性
4. 找到中指

#include 
    
     int Odd(int n);void orderIntegerArray(int *p, int n);double findMedianSortedArrays(int *numS1, int n1, int *sumS2, int n2);int main(void){    int nums1[9] = {3, 13, 7, 5, 21, 23, 23, 40, 23};    int nums2[5] = {14, 12, 56, 23, 29};    printf("The median is %lf\n", findMedianSortedArrays(nums1, 9, nums2, 5));    return 0;}double findMedianSortedArrays(int *numS1, int n1, int *numS2, int n2){    int *p, *q;    p = numS1;    q = numS2;    int newArray[n1 + n2];    int *k;    k = newArray;    for (int i = 0; i < n1; i++)        *(k + i) = *(p + i);    for (int i = n1; i < n1 + n2; i++)        *(k + i) = *(q + i - n1);    orderIntegerArray(newArray, n1 + n2);    for (int i = 0; i < n1 + n2; i++)        printf("%d---", *(k + i));    putchar('\n');    double median;    int n3 = n1 + n2;    if (Odd(n3))        median = newArray[(n3 + 1) / 2 - 1];    else        median = (newArray[n3 / 2 - 1] + newArray[n3 / 2 + 1 - 1]) / 2;    return median;}void orderIntegerArray(int *p, int n){    int temp;    for (int i = 0; i < n; i++)        for (int j = i + 1; j < n; j++)            if (*(p + i) > *(p + j))            {                temp = *(p + i);                *(p + i) = *(p + j);                *(p + j) = temp;            }}int Odd(int n){    if (n % 2 == 0)        return 0;// even    else        return 1; // odd}
    

输出为：

3----5----7----12----13----14----21----23----23----23----23----29----40----56----