Binary String Matching

题目描述：

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.

接到这个题首先想到的就是在学习数字逻辑和编译原理时用过的状态图。但是这里也可以用一个更加简单的方式，判断字串来解决。

import java.util.*;/** * 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 * @author sdu20 * */public class BinaryString {public static void main(String[] args) {// TODO Auto-generated method stubScanner in = new Scanner(System.in);try{String str = in.nextLine();int num = Integer.valueOf(str).intValue();int[] con = new int[num];for(int i = 0;i<num;i++){String sub = in.nextLine();String par = in.nextLine();con[i] = contains(par,sub);}for(int i = 0;i<num;i++)System.out.println(con[i]);}catch(Exception e){e.printStackTrace();}}/** * 检测a中有多少个b * @param a * @param b * @return */public static int contains(String a,String b){if(a.equals(b))return 1;if(a.length()<=b.length())return 0;int numbers = 0;for(int i = 0;i<a.length()-b.length();i++){if(a.substring(i, i+b.length()).equals(b))numbers++;}return numbers;}}

运行截图如下：