实验11-1-8 查找子串(20 分)

实验11-1-8 查找子串(20 分)

本题要求实现一个字符串查找的简单函数。

函数接口定义:

1
char *search( char *s, char *t );

函数search在字符串s中查找子串t,返回子串t在s中的首地址。若未找到,则返回NULL。

裁判测试程序样例:

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#include <stdio.h>
#define MAXS 30

char *search(char *s, char *t);
void ReadString( char s[] ); /* 裁判提供,细节不表 */

int main()
{
char s[MAXS], t[MAXS], *pos;

ReadString(s);
ReadString(t);
pos = search(s, t);
if ( pos != NULL )
printf("%d\n", pos - s);
else
printf("-1\n");

return 0;
}

/* 你的代码将被嵌在这里 */

输入样例1:

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2
The C Programming Language
ram

输出样例1:

1
10

输入样例2:

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2
The C Programming Language
bored

输出样例2:

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-1
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#include <stdio.h>
#define MAXS 30

char *search(char *s, char *t);
void ReadString( char s[] ); /* 裁判提供,细节不表 */

int main()
{
char s[MAXS], t[MAXS], *pos;

ReadString(s);
ReadString(t);
pos = search(s, t);
if ( pos != NULL )
printf("%d\n", pos - s);
else
printf("-1\n");

return 0;
}
void ReadString(char s[]){
gets(s);
}
char *search(char *s, char *t)
{
int i, j, sLen, tLen;
for (i=0; s[i]!='\0'; i++);
sLen=i;
for (j=0; t[j]!='\0'; j++);
tLen=j;
i=j=0;
while (i < sLen && j < tLen)
{
if (s[i] == t[j])
{
i++;
j++;
}
else
{
i = i - j + 1;
j = 0;
}
}
if (j == tLen)
return &s[i - j];
else
return NULL;
}
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