Binary String Matching二進位制字串匹配
阿新 • • 發佈:2018-12-30
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
輸入
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.
輸出
For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
package suanfa; import java.util.Scanner; public class Fiveth { public static void main(String[] args){ Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int result[]=new int[n]; String a,b; for(int j=0;j<n;j++) { a = sc.next(); b = sc.next(); int sum=0; for(int i=0;i<b.length()-a.length()+1;i++){ if(a.equals(b.substring(i,i+a.length()))){ sum++; } } result[j]=sum; } for(int i=0;i<n;i++) { System.out.println(result[i]); } }}