Java按顺序检查数组是否包含指定子数组问题求助

Hey there! Let's work through this problem where you need to check if your input array arr2 appears in arr1 in the correct order—either as a consecutive subarray (which matches your example where [21,19] returns false) or just as a subsequence (order matters but elements don't have to be adjacent). Here's how to fix your code:

First, Let's Clarify the Two Scenarios #

Based on your examples:

• When arr1 = [45,21,1,19,8,90,21,2] and arr2 = [21,1,19], you want true (this works as both a consecutive subarray and subsequence)
• When arr1 = [45,21,1,19,8,90,21,2] and arr2 = [21,19], you want false—this means you're looking for consecutive elements, not just ordered ones.

Solution 1: Check for Consecutive Subarray #

This method verifies if arr2 exists as a continuous block in arr1 with the exact order.

import java.util.Random;
import java.util.Scanner;

public class Main {
    private static int[] array1;

    // Initialize arr1 with random numbers
    public static int[] list() {
        array1 = new int[10];
        Random random = new Random();
        for (int i = 0; i < array1.length; i++) {
            array1[i] = random.nextInt(100); // Generate random numbers between 0-99
        }
        return array1;
    }

    // Check if arr2 is a consecutive subarray of arr1
    public static boolean isConsecutiveSubarray(int[] arr2) {
        // Edge cases: empty arr2 is considered a match, or adjust if needed
        if (arr2 == null || arr2.length == 0) {
            return true;
        }
        // If arr2 is longer than arr1, it can't be a subarray
        if (arr2.length > array1.length) {
            return false;
        }

        // Iterate through all possible starting positions in arr1
        for (int i = 0; i <= array1.length - arr2.length; i++) {
            boolean isMatch = true;
            // Compare each element of arr2 to the consecutive elements in arr1
            for (int j = 0; j < arr2.length; j++) {
                if (array1[i + j] != arr2[j]) {
                    isMatch = false;
                    break;
                }
            }
            if (isMatch) {
                return true;
            }
        }
        // No matching consecutive block found
        return false;
    }

    // Get user input to create arr2
    public static int[] getUserInputArray() {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter the number of elements in arr2: ");
        int size = scanner.nextInt();
        int[] arr2 = new int[size];
        System.out.println("Enter the elements separated by spaces:");
        for (int i = 0; i < size; i++) {
            arr2[i] = scanner.nextInt();
        }
        scanner.close();
        return arr2;
    }

    public static void main(String[] args) {
        // Initialize the random array arr1
        array1 = list();
        System.out.println("Generated arr1: ");
        for (int num : array1) {
            System.out.print(num + " ");
        }
        System.out.println();

        // Get user's arr2
        int[] arr2 = getUserInputArray();

        // Check and print result
        boolean result = isConsecutiveSubarray(arr2);
        System.out.println("Does arr1 contain arr2 as a consecutive subarray? " + result);
    }
}

Solution 2: Check for Subsequence (Order Matters, No Need for Consecutive) #

If you ever need to allow arr2 elements to appear in order but not necessarily next to each other (e.g., [21,19] would return true for your sample arr1), use this method instead:

// Check if arr2 is a subsequence of arr1
public static boolean isSubsequence(int[] arr2) {
    if (arr2 == null || arr2.length == 0) {
        return true;
    }
    int arr2Pointer = 0;
    // Traverse arr1 and look for arr2 elements in order
    for (int num : array1) {
        if (num == arr2[arr2Pointer]) {
            arr2Pointer++;
            // If we've found all elements of arr2, return true
            if (arr2Pointer == arr2.length) {
                return true;
            }
        }
    }
    // Return true only if all elements of arr2 were found in order
    return arr2Pointer == arr2.length;
}

You can replace isConsecutiveSubarray with isSubsequence in the main method to use this logic instead.

内容的提问来源于stack exchange，提问作者Ali Göktaş