寻找子串匹配的个数

描述

Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

输入

The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 100. And it is guaranteed that B is always longer than A.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011 

样例输出

3
0
3 

import java.util.Scanner;

public class Main05 {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();            //测试数据的组数
        while(n != 0)
        {
            String A = scan.next();
            String B = scan.next();
            String temp = "";
            int count = 0;
            if(A.length() > B.length())
                System.out.println(count);
            else
            {
                for(int i = 0;i <= B.length()-A.length();i++)
                {
                    temp = B.substring(i, i+A.length());//注意字符串截取时最后一个下标不包含
                    if(temp.equals(A))
                        count++;
                }
                System.out.println(count);    
            }
            n--;
        }
    }
}



很简单，也没有用什么匹配算法，java里面提供了截取字串的方法，直接去字串就行，提交了2次才过，注意的是截取字串是从start（下标包括）到end（下标不包括）