NYOJ之Binary String Matching

Binary String Matching

时间限制：3000 ms  |  内存限制：65535 KB
难度：3

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

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AC代码：

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        
        Scanner sc=new Scanner(System.in);
        int n=Integer.parseInt(sc.nextLine());
        
        while(n-->0){
            String a=sc.nextLine();
            String b=sc.nextLine();
            int ans=solve(a,b);
            System.out.println(ans);
        }
        
    }
    
    public static int solve(String a,String b){
        int res=0;
        for(int i=0;i<b.length();i++){
            int j=0;
            for(j=0;j<a.length();j++){
                if((i+j>=b.length()) || (a.charAt(j)!=b.charAt(i+j))) break;
            }
            if(j==a.length()) res++;
        }
        return res;
    }
    
}



另一种AC思路：（利用了内建的方法，不重复造轮子）

import java.util.Scanner;

public class Main {
    
    public static void main(String[] args) {
        
        Scanner sc=new Scanner(System.in);
        int n=Integer.parseInt(sc.nextLine());
        
        while(n-->0){
            String a=sc.nextLine();
            String b=sc.nextLine();
            int ans=solve(a,b);
            System.out.println(ans);
        }
        
    }
    
    
    public static int solve(String a,String b){
        int res=0;
        while(a.length()<=b.length()){
            if(b.indexOf(a)==0) res++;
            b=b.substring(1,b.length());
        }
        return res;
    }
    
}

 

 

题目：http://acm.nyist.net/JudgeOnline/problem.php?pid=5