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.ArrayList;import java.util.HashMap;import java.util.List;import java.util.Map;import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();List<Object> list = new ArrayList<Object>();for (int i = 0; i < n; i++) {Map<String, String> map = new HashMap<String, String>();String str1 = sc.next();String str2 = sc.next();map.put("str1", str1);map.put("str2", str2);list.add(map);}print(list);}private static void print(List<Object> list) {for (int i = 0; i < list.size(); i++) {@SuppressWarnings("unchecked")Map<String, String> map = ((Map<String, String>) list.get(i));String str1 = (String) map.get("str1");String str2 = (String) map.get("str2");match(str1, str2);}}static void match(String str1, String str2) {int i, count = 0;i = str2.indexOf(str1);// strB在strA中首次出现的位置while (i >= 0) {count++;i = str2.indexOf(str1, i + 1);}System.out.println(count);}}