下面的函式統計子字串substr在字串str中出現的次數,如果substr在str中不出現,則返回值0
阿新 • • 發佈:2019-02-12
c++
#include<iostream>
#include<string>
using namespace std;
int str_count(string substr, string str)
{
int times=0, num=0,i=0;
int a=substr.length();
int b=str.length();
if(a>b)
return 0;
else
{
for(int i=0;i<b;i++)
{
if(str[i]==substr[0 ])
for(int j=0;j<a;j++)
{
if(str[i+j]==substr[j])
{
num++;
continue;
}
else
break ;
}
}
times=num/a;
return times;
}
}
int main()
{
string str;
cout<<"plz input str:"<<endl;
cin>>str;
string substr;
cout<<"plz input substr:"<<endl;
cin>>substr;
cout<<"the repeated is " <<str_count(substr,str)<<" times"<<endl;
while(1);
return 0;
}
java: 暴力方法
//substr在str中出現的次數
public class Substr {
public static void main(String[] args) {
String str = "ajavabjavacjavadjavaejavafjava";
String substr = "java";
int len = substr.length();
int index ;
int count = 0;
while( (index=(str.indexOf(substr)) )!= -1) {
count ++;
str = str.substring(index+len);
}
System.out.println(count);
}
}