Read ahead in an array to predict later outcomes i

问题:

Basically what I'm trying to ask is, is there anyway to read ahead in an array so that you could create a 'case' for it.

For ex: you array has only integers such as: 0 0 0 0 4 0 1 0 0 0 0 0 0 0 0 0 0 0 0 3

and what you want to try to do is create a coutdown till the next non zero number. Basically display the countdown. Is there any way to do this?

回答1:

This code:

#include <stdio.h>

int main(void)
{
    int array[] = { 0, 0, 0, 0, 4, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, };

    int a_size  = sizeof(array) / sizeof(array[0]);

    int i = 0;
    while (i < a_size)
    {
        if (array[i] == 0)
        {
            int j;
            for (j = i; j < a_size; j++)
                if (array[j] != 0)
                    break;
            printf("%d\n", j - i);
            i = j;
        }
        else
            i++;
    }
    return 0;
}

produces this output:

4
1
12

If that's what you want, that's roughly what you need. If it is not what you want, you need to explain more clearly what it is that you want.

---

Revised code for revised expected output:

#include <stdio.h>

int main(void)
{
    int array[] = { 0, 0, 0, 0, 4, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, };

    int a_size  = sizeof(array) / sizeof(array[0]);

    int i = 0;
    while (i < a_size)
    {
        if (array[i] == 0)
        {
            int j;
            for (j = i; j < a_size; j++)
                if (array[j] != 0)
                    break;
            int k = j - i;
            while (k > 0)
                printf(" %d", k--);
            i = j;
        }
        else
        {
            printf(" '");
            i++;
        }
    }
    putchar('\n');
    return 0;
}

Revised output:

 4 3 2 1 ' 1 ' 12 11 10 9 8 7 6 5 4 3 2 1 '