寻找子串匹配的个数

最新推荐文章于 2023-04-01 14:11:06 发布

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最新推荐文章于 2023-04-01 14:11:06 发布

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文章标签： java

本文介绍了一个简单的Java程序，用于解决两个仅由'0'和'1'组成的字符串A和B的问题，任务是计算字符串A作为子串在B中出现的次数。通过逐个检查B中的子串是否与A相等来实现。

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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) <= 100. 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 Main05 {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int n = scan.nextInt(); //测试数据的组数
while(n != 0)
{
String A = scan.next();
String B = scan.next();
String temp = "";
int count = 0;
if(A.length() > B.length())
System.out.println(count);
else
{
for(int i = 0;i <= B.length()-A.length();i++)
{
temp = B.substring(i, i+A.length());//注意字符串截取时最后一个下标不包含
if(temp.equals(A))
count++;
}
System.out.println(count);
}
n--;
}
}
}

很简单，也没有用什么匹配算法，java里面提供了截取字串的方法，直接去字串就行，提交了2次才过，注意的是截取字串是从start（下标包括）到end（下标不包括）