java--二进制字符串匹配的问题

最新推荐文章于 2021-03-05 04:04:43 发布

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#二进制字符串匹配 #JAVA

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本文介绍了一个简单的字符串匹配问题，即计算一个由'0'和'1'组成的较短字符串A在另一个较长字符串B中作为子串出现的次数。通过一个Java程序实现了解决方案，并给出了具体的示例输入和输出。

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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

题目大意：输入两个字符串A，B，只包含0或者1，有多少次A出现在B的字符串中，例如B是1001110110，A是11，有3次A出现在B中

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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();//n=3
		
		String str;
		String pattern;
		//循环3次
		while(n > 0){
			pattern = sc.next();//11
			str = sc.next();//1001110110
			
			int index = 0;//记录pattern在str中出现的索引
			int count = 0;//记录出现的次数，默认=0
			
			while(index != -1){
				index = str.indexOf(pattern);//index = 3
				if(index != -1){
					//说明存在pattern
					count ++ ;//count = 1
					str = str.substring(index+1);//此时str = 110110
				}
			}
			
			System.out.println(count);
			
			
			n--;
		}
	}
}