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*）
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;
}
}