Read ahead in an array to predict later outcomes in C [closed]

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 '

来源：https://stackoverflow.com/questions/20167106/read-ahead-in-an-array-to-predict-later-outcomes-in-c

标签

arrays

countdown

pic32