TCPL——练习3.3

　　TCPL练习3－3的题目是这样的：函数expand(s1, s2)，将字符串中s1中类似a-z的速记扩展为：abcd...xyz，并能处理大小写，并可以处理a-b-c-d, a-z0-9, -a-z这样的情况，前导的-字符照样排印。

　　我记得这道题在豆瓣上看朋友用PY写得很简洁，用C写的代码看起来比较累赘，可能也是我想得比较复杂。代码：

 

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

#define MAXSIZE 100
#define MINSIZE 50

static int is_iterable(int c)
{
    if (islower(c) || isupper(c) || isdigit(c) ) {
        return 1;
    }
    return 0;
}

int expand(const char *s1, char *s2)
{
    int i, j;
    int b, e;
    int count = 0;
    int s2_len = 0;

    for (i = 0; s1[i] == '-'; ++i) {
        s2[s2_len++] = '-';
    }
    for ( ; s1[i] != '\0'; ++i) {
        if (is_iterable(s1[i]) ) {
            if (s1[i+1] == '-') {
                b = s1[i];
                e = s1[i + 2];
                for (j = 0, count = e - b + 1; j < count; ++b, ++j) {
                    if (s2[s2_len-1] != b) {
                        s2[s2_len++] = b;
                    }
                }
            } else {
                s2[s2_len++] = s1[i];
            }
        } else {
            s2[s2_len++] = s1[i];
        }
　　　　　i += s1[i+3] == '-' ? 1 : 2; 
    }
    return s2_len;
}

int main(void)
{
    int len;
    char s1[MINSIZE] = {0};
    char s2[MAXSIZE] = {0};

    while (1) {
        printf("input string(a-z,0-9,.., 'quit' to exit):\n");

        memset(s1, 0x00, sizeof(s1));
        fgets(s1, MINSIZE, stdin);
        s1[strlen(s1) - 1] = '\0';

        if (memcmp(s1, "quit", 4) == 0)
            break;

        memset(s2, 0x00, sizeof(s2));
        len = expand(s1, s2);
        printf("====> %s\t\tlen = %d\n", s2, len);
    }

    return 0;
}

执行结果如下：

 

 

ken@Linux:~/TCPL/chap3$ ./exe3_3 
input string(a-z,0-9,.., 'quit' to exit):
a-z
====> abcdefghijklmnopqrstuvwxyz        len = 26
input string(a-z,0-9,.., 'quit' to exit):
--a-b-c0-9
====> --abc0123456789        len = 15
input string(a-z,0-9,.., 'quit' to exit):
0-9
====> 0123456789        len = 10
input string(a-z,0-9,.., 'quit' to exit):
a-b-c-d-e-f-uc-u0-9
====> abcdefghijklmnopqrstucdefghijklmnopqrstu0123456789        len = 50
input string(a-z,0-9,.., 'quit' to exit):
quit

 

PS：刚才看了TCAB，里面的答案写得很简洁，确实是我想得复杂了，不应该把情况分得那么细。整体的思路还是一样的，重新看了自己的代码，发现有好些地方是可以省略或者归并的。下面的TCAB里面的代码，学习学习：

 

void expand(char s1[], char s2[])
{
    char c;
    int i, j;

    i = j = 0;
    while ((c = s1[i++]) != '\0') {
        if (s1[i] == '-' && s1[i+1] >= c) {
            i++;
            while (c < s1[i])
                s2[j++] = c++;
        } else
            s2[j++] = c;
    }
    s2[j] = '\0';
}