指针求两个有序数组的第一个相同元素

first common element

// 给定两个递增有序表，输出两表第一个相同的元素

#include <stdio.h>
#include <string.h>

int *fst_common_val(int *a,int an, int *b, int bn)
{
    int *ptra, *ptrb;
    ptra = a; ptrb = b;
    while(ptra < a+an && ptrb < b + bn)
    {
        if(*ptra < *ptrb) ptra ++;
        else if(*ptra > *ptrb) ptrb++;
        else
            return ptra;
    }
    return ptra;
}

int main()
{
    int *a[] = {1,3,5,7,9,11,13,17,19};
    int *b[] = {2,4,6,8,11,56,};
    int *value;
    int (*func)();
    func = fst_common_val;
    printf("the elements of a are: \n");
    for(int i=0; i<sizeof(a)/sizeof(a[0]); i++)
        printf("%d ", *(a+i));

    printf("\nthe elements of b are: \n");
    for(int i=0; i<sizeof(b)/sizeof(b[1]); i++)
        printf("%d ", *(b+i));

    value = (*func)(a,sizeof(a)/sizeof(a[0]), b, sizeof(b)/ sizeof(b[0]));
    printf("\nthe fist common element is : %d", *value);
    return 0;
}

代码是强行用指针的，为了练习指针的使用，不是最优解

来源：CSDN

作者：ever_promise

链接：https://blog.csdn.net/ever_promise/article/details/104268866