ZZULIOJ 1147: Finding subarrays, Java

ZZULIOJ 1147: Finding subarrays, Java

Question description

Given two integer arrays, array a has n elements, array b has m elements, 1<=m<=n<100, please check whether array b is a subarray of array a. If starting from an element a[i] of array a, there are b[0]=a[i], b[1]=a[i+1],…, b[m]=a[i+m] , then array b is said to be a subarray of array a.

enter

The first line of input is two integers n and m; the second line is the n integers of array a; the third line is m integers of array b, with spaces separating each data.

output

The output occupies one line. If b is a subarray of a, output the position i of the subarray. Note that the subscript starts from 0; otherwise, output No Answer.

Sample inputCopy

8 3
3 2 6 7 8 3 2 5
3 2 5

Sample outputCopy

5

import java.io.*;

public class Main {
    
    
    public static void main(String[] args) throws IOException {
    
    
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        String[] str = bf.readLine().split(" ");
        int n = Integer.parseInt(str[0]);
        int m = Integer.parseInt(str[1]);
        int[] a = new int[n], b = new int[m];
        str = bf.readLine().split(" ");
        for (int i = 0; i < n; i++) {
    
    
            a[i] = Integer.parseInt(str[i]);
        }
        str = bf.readLine().split(" ");
        for (int i = 0; i < m; i++) {
    
    
            b[i] = Integer.parseInt(str[i]);
        }
        boolean ok = false;
        for (int i = 0; i < n - m + 1; i++) {
    
    
            int j;
            for (j = 0; j < m; j++) {
    
    
                if (a[i + j] != b[j]) {
    
    
                    break;
                }
            }
            if (j == m) {
    
    
                ok = true;
                bw.write(i + "\n");
                break;
            }
        }
        if (!ok) bw.write("No Answer\n");
        bw.close();
    }
}