自己写了一个计算字符串中子串出现的次数的程序，不知道大家有没有更好的算法拿来分享下！
package com.niit.demo;

import java.util.Scanner;

public class Str
{
    private static int sum;
    public static void main(String[] args) {
        System.out.println("请输入字符串：");
        Scanner sn=new Scanner(System.in);
        String sur=sn.next();
        System.out.println("请输入您要统计的子字符串：");
        String dest=sn.next();
        System.out.println("正在统计子字符串出现的次数...");

        Str s=new Str();
        int times=s.fun(sur,dest);
        System.out.println("出现的次数为："+times);
    }
   
    public int fun(String sur,String dest)
    {
        int index=sur.indexOf(dest);
        int last=sur.lastIndexOf(dest);
        if(index!=-1 && last!=-1 )
        {
            if(index==last)
            {
                sum=1;
            }
            else
            {
                String temp = sur.substring(index, sur.length());
                while (true) {
                    int x = temp.indexOf(dest);
                    if (x == -1) {
                        break;
                    } else {
                        sum++;
                        temp = temp.substring(x+dest.length(), temp.length());
                    }
                }
            }
        }
        return sum;
    }   
}

请输入字符串：
010101010
请输入您要统计的子字符串：
010
正在统计子字符串出现的次数...
出现的次数为：2

哇 ，还真有bug来！
多谢版主，我再改改！

import java.util.Scanner;
public class ArraysSubstring {
    public static void main(String[] args) {
        System.out.println("请输入字符串：");
        Scanner sn=new Scanner(System.in);
        String sur=sn.next();
        System.out.println("请输入您要统计的子字符串：");
        String dest=sn.next();
        System.out.println("正在统计子字符串出现的次数...");

        ArraysSubstring s=new ArraysSubstring();
        int times=s.fun(sur,dest);
        System.out.println("出现的次数为："+times);
    }
   
    public int fun(String sur,String dest)
    {
        int sum = 0;
        int index=sur.indexOf(dest);
        while(index!=-1){
            sum+=1;
            index = sur.indexOf(dest, index+1);
        }     
        return sum;
    }    

}