Binary String Matching Java实现

题目如下：

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

31110011101101011100100100100011010110100010101011 

样例输出

303 

算法分析：

利用Java自带的substring(beginningindex，endindex)提取字符串B中的字串，利用match函数比较，若相同则计数加1.

代码如下：

package binarymatch;import java.util.*;public class Main {public static void main(String[] args){Scanner input = new Scanner(System.in);int num = 0;//System.out.println("输入num");num = input.nextInt();String[] A = new String[num];String[] B = new String[num];for(int index = 0; index < num ; index++){//System.out.println("输入A");A[index] = input.next();//System.out.println("输入B");B[index] = input.next();}for(int index = 0; index < num ; index++){System.out.println(binary_string_match(A[index],B[index]));}}public static int binary_string_match(String A, String B){int ret = 0;int  lengtha = A.length();int  lengthb = B.length();for(int index = 0; index < (lengthb-lengtha+1); index++){String sub = new String();sub = B.substring(index, index+lengtha);if(sub.matches(A))ret++;}return ret;}}