Topic: Implement a string search program.
Requirement: Given a string s=princess-connect-redive, let the user find connect and judge whether connect is in the string s.
Do not use built-in string functions, please write your own program to implement a string search algorithm.
Problem-solving ideas: write a string search function
Look at the main function first
int main()
{
char s[]="princess-connect-redive";
char str0[]="connect";
int res = strcompare(str0,s);
printf("大家好，这是字符串查找程序,如下：\n");
if(res)
printf("%s在%s中\n\n",str0,s);
else
printf("%s不在%s中\n\n",str0,s);
while(1) {
strinput();}
return 0;
}
Key functions:
1. The function strlength to obtain the length of a string
/************************
函数：int strlength(char *s)
功能：返回输入字符串的长度
参数：字符串 char *s
返回值：int整型
*************************/
int strlength(char *s)
{
int count = 0;;
for(int i=0;i<100;i++)
{
if(s[i]=='\0')
break;
else
count++;
}
return count;
}
2. String search and comparison function strcompare
/************************
函数：int strcompare(char *s)
功能：输入字符串str1,str2，判断str1是否在str2中，
是则返回1，否则返回0
参数：字符串 char *str1,char *str2
返回值：int整型（1或者0）
*************************/
int strcompare(char *str1,char *str2)
{
int res,n,m;
n = strlength(str1);
m = strlength(str2);
if(n>m)
res = 0;
else if(str1[0]=='\0')
res = 0;
else
{
int count = 0,i = 0;
for(int j=0;j<m;j++)
{
if(str1[i]!='\0' || str2[j]!='\0')
break;
A: if(str1[i] == str2[j])
{
i++;
j++;
count++;
goto A;
}
else
{
if(count==n)
return res=1;
else
{
if(j==m)
{
res = 0;
break;
}
else
{
i = 0;
count = 0;
}
}
}
}
}
return res;
}
3. The string input function strinput
/************************
函数：int strinput(void)
功能：从键盘输入两个字符串，判断要查找的字符串是否在
被查找的字符串之中，并输出比较结果
参数：无
返回值：无
*************************/
void strinput(void)
{
char str1[100],str2[100];
printf("请输入bei查找的字符串:");
gets(str2);
printf("请输入yao查找的字符串:");
gets(str1);
int res = strcompare(str1,str2);
if(res)
printf("查找结果：%s在%s中\n\n",str1,str2);
else
printf("查找结果：%s不在%s中\n\n",str1,str2);
}
The complete code is as follows:
//字符串查找
#include <stdio.h>
int strlength(char *s);
int strcompare(char *str1,char *str2);
void strinput(void);
int main()
{
char s[]="princess-connect-redive";
char str0[]="connect";
int res = strcompare(str0,s);
printf("大家好，这是字符串查找程序,如下：\n");
if(res)
printf("%s在%s中\n\n",str0,s);
else
printf("%s不在%s中\n\n",str0,s);
B: strinput();
goto B;
return 0;
}
//返回字符串的长度，要求字符串的长度不能超过100
int strlength(char *s)
{
int count = 0;;
for(int i=0;i<100;i++)
{
if(s[i]=='\0')
break;
else
count++;
}
return count;
}
//查找字符串str1是否在str2中，是则返回1，否则返回0
int strcompare(char *str1,char *str2)
{
int res,n,m;
n = strlength(str1);
m = strlength(str2);
if(n>m)
res = 0;
else if(str1[0]=='\0')
res = 0;
else
{
int count = 0,i = 0;
for(int j=0;j<m;j++)
{
if(str1[i]!='\0' || str2[j]!='\0')
break; //少了这个if检测最后一个字符就会出错！
A: if(str1[i] == str2[j])
{
i++;
j++;
count++;
goto A;
}
else
{
if(count==n)
return res=1;
else
{
if(j==m)
{
res = 0;
break;
}
else
{
i = 0;
count = 0; //count必须要复位!
}
}
}
}
}
return res;
}
void strinput(void)
{
char str1[100],str2[100];
printf("请输入bei查找的字符串:");
gets(str2);
printf("请输入yao查找的字符串:");
gets(str1);
int res = strcompare(str1,str2);
// printf("被查找的字符串的长度是%d\n",strlength(str2));
// printf("要查找的字符串的长度是%d\n",strlength(str1));
if(res)
printf("查找结果：%s在%s中\n\n",str1,str2);
else
printf("查找结果：%s不在%s中\n\n",str1,str2);
}
The operation effect is as follows:
