The C Programming Language 练习题2-4

题目

squeeze(s1, s2)，将字符串s1 中任何与字符串s2 中字符匹配的字符都删除。

题目分析

先将s2中每个字符串拿出来在s1中寻找，然后生成不包含此字符的新的s1字符串，然后再找下一个。

编程实现

为了方便定位问题，加的打印信息较多。。。。（其实已经删掉许多打印了）

#include <stdio.h>
#define MAXLINE 1000

void squeeze(char s1[], char s2[]);

int main()
{
int i;
char c, sfirst[MAXLINE], ssecond[MAXLINE];

i = 0;
printf("Please input string1:");
while ((c = getchar()) != '\n')
sfirst[i++] = c;
sfirst[i] = '\0';
printf("The first string is:");
i = 0;
while (sfirst[i] != '\0')
printf("%c", sfirst[i++]);
printf("\n");

i = 0;
printf("Please input string2:");
while ((c = getchar()) != '\n')
ssecond[i++] = c;
ssecond[i] = '\0';
printf("The second string is:");
i = 0;
while (ssecond[i] != '\0')
printf("%c", ssecond[i++]);
printf("\n");

squeeze(sfirst, ssecond);
i = 0;
while (sfirst[i] != '\0')
{
printf("%c",sfirst[i]);
i++;
}
}

void squeeze(char s1[], char s2[])
{
int m, n, l;

m = n = l = 0;
for (n = 0; s2
!= '\0'; n++)
{
for (m = l = 0; s1[m] != '\0'; m++)
if (s1[m] != s2
)
s1[l++] = s1[m];
s1[l] = '\0';
}
}