Binary String Matching

最新推荐文章于 2025-09-08 22:39:06 发布

转载最新推荐文章于 2025-09-08 22:39:06 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/zqxLonely/p/4129540.html

文章标签：

#java

本文介绍了一个简单的Java程序，用于解决给定两个仅由'0'和'1'组成的字符串A和B时，如何计算字符串A作为子串在B中出现的次数。通过使用Java内置的indexOf方法进行遍历和匹配，程序能够高效地找到所有匹配项并计算总数。

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) <= 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

 1 import java.util.Scanner;
 2 
 3 public class Main {
 4     public static void main(String[] args) {
 5         Scanner scanner=new Scanner(System.in);
 6         int T;
 7         String A;
 8         String B;
 9         int startPoint;
10         int count;
11         
12         T=scanner.nextInt();
13         while(T!=0){
14             T--;
15             startPoint=0;
16             count=0;
17             
18             A=scanner.next();
19             B=scanner.next();
20             
21             while(true){
22                 startPoint=B.indexOf(A,startPoint);
23                 
24                 if(startPoint==-1)
25                     break;
26                 
27                 count++;
28                 startPoint++;
29             }
30             
31             System.out.println(count);
32         }
33     }
34 }