Binary String Matching

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) <= 1000. 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 No5 {

    private int stringCheck(String aStr, String bStr) {
        try {
            StringBuffer aSB = new StringBuffer(aStr);
            StringBuffer bSB = new StringBuffer(bStr);
　　　　　　  // 窗口滑动次数
            int step = bStr.length() - aStr.length() + 1;
　　　　　　　// 模拟滑动窗口
            int window = aStr.length();
            int equalCount = 0;

            for (int i = 0; i < step; i++) {
                int result = ((String) bStr.subSequence(i, i + window))
                        .compareTo(aStr);
                if (result == 0) {
                    equalCount++;
                }
            }
            return equalCount;
        } catch (RuntimeException e) {

        }
        return -1;
    }

    public static void main(String args[]) throws Exception {
        try {
            Scanner cin = new Scanner(System.in);
            int num = cin.nextInt();
            int[] result = new int[num];
            No5 no5 = new No5();

            for (int i = 0; i < num; i++) {
                String aStr = cin.next();
                String bStr = cin.next();
                result[i] = no5.stringCheck(aStr, bStr);
            }

            for (int i = 0; i < num; i++) {
                System.out.println(result[i]);
            }
        } catch (RuntimeException e) {

        }
    }

}