Binary String Matching

最新推荐文章于 2019-03-12 21:23:39 发布

原创最新推荐文章于 2019-03-12 21:23:39 发布 · 10w+ 阅读

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文章标签：

#ACM #Java

本文介绍了一种解决二进制字符串匹配问题的算法，即如何确定一个仅由'0'和'1'组成的字符串A作为子串在另一个较长的二进制字符串B中出现的次数。通过遍历字符串B并检查所有可能的子串是否与A相等来实现，适用于字符串长度不超过特定限制的情况。

Binary String Matching

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

Java实现

	private static int match(String a, String b) {
		int result = 0;
		int al = a.length();
		int bl = b.length();
		for(int i = 0; i < (bl - al + 1); i++){
			String sub = b.substring(i, i + al);
			if(sub.matches(a)){
				result++;
			}
		}
		return result;
	}
	
	private static void binaryStringMatch() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		String[] a = new String[n];
		String[] b = new String[n];
		for(int i = 0; i < n; i++){
			a[i] = sc.next();
			b[i] = sc.next();
		}
		for(int i = 0; i < n; i++){
			System.out.println(match(a[i], b[i]));
		}
	}