java--二進位制字串匹配的問題
阿新 • • 發佈:2019-01-23
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
題目大意:輸入兩個字串A,B,只包含0或者1,有多少次A出現在B的字串中,例如B是1001110110,A是11,有3次A出現在B中
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. 樣例輸入3 11 1001110110 101 110010010010001 1010 110100010101011
樣例輸出
3 0 3
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; //迴圈3次 while(n > 0){ pattern = sc.next();//11 str = sc.next();//1001110110 int index = 0;//記錄pattern在str中出現的索引 int count = 0;//記錄出現的次數,預設=0 while(index != -1){ index = str.indexOf(pattern);//index = 3 if(index != -1){ //說明存在pattern count ++ ;//count = 1 str = str.substring(index+1);//此時str = 110110 } } System.out.println(count); n--; } } }