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java--二进制字符串匹配的问题
2013-06-17 14:57:35      个评论      
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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

 

[java]
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--; 
        } 
    } 

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--;
  }
 }
}


 

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