Java tutorial: Implement digital sequence filtering and specified length output

This tutorial aims to guide developers how to use the Java programming language to generate a number sequence starting from 1 based on the specified number `n` input by the user. This sequence will strictly exclude all integers containing the number '1' and ensure that the final number of output numbers is exactly `n`. The article will introduce two core implementation methods: a simple solution based on string conversion and an efficient solution based on mathematical operations, and provide detailed code examples and logical analysis.

Requirements analysis and problem overview

In programming practice, we often encounter scenarios where we need to filter and process digital sequences. This tutorial focuses on a specific problem: the user inputs an integer n, and the program needs to find the number in increments starting from 1. If a number contains the number '1' (such as 1, 10, 121, etc.), then skip the number; otherwise, add it to the output sequence. This process continues until the output sequence contains n numbers that meet the conditions. The final number of numbers output must exactly match the value of n entered by the user.

For example, if the user enters 22, the program should output a sequence of 22 numbers, none of which contain '1', in ascending order. The output may be: 2, 3, 4, 5, 6, 7, 8, 9, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35.

Implementation method one: String conversion method

The core idea of ​​this method is to convert each integer to be checked into a string, and then use the contains() method of the string to determine whether it contains the character '1'.

logical steps

1. Get user input: Use the Scanner class to get the integer n input by the user.
2. Initialize counter: Set a counter count to record the number of numbers that have been output that meet the conditions. The initial value is 0.
3. Loop iteration: Use a for loop, starting from i = 1 and incrementing, and the loop condition is count
4. Checking numbers: Inside the loop, convert the current number i to a string String.valueOf(i).
5. Determine whether it contains '1': Use String.valueOf(i).contains("1") to determine whether the string contains '1'.
6. Skip or output:
   • If '1' is included, use the continue statement to skip the remainder of the current loop and go directly to the next iteration.
   • If it does not contain '1', print the number i and increase count by 1.
7. Close the resource: Close the Scanner object when the program ends.

Sample code

 import java.util.Scanner;

public class NumberFilterString {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Please enter an integer n:");
        int num = sc.nextInt(); // Get the n entered by the user

        int count = 0; // Initialize the counter of the output numbers System.out.println("The qualified number sequence is as follows:");

        // Loop to find and output n qualified numbers for (int i = 1; count <h4> Advantages and Disadvantages</h4>

• Advantages: The code is concise and easy to understand, and it uses Java's built-in string processing function to achieve intuitive implementation.
• Disadvantages: Each iteration requires an integer-to-string conversion, which may cause some performance overhead when dealing with large numbers.

Implementation method two: mathematical operation method

In order to avoid the performance loss caused by string conversion, we can use pure mathematical operations to check whether a number contains '1'. This method checks each bit of the number by repeatedly taking modulo and dividing.

logical steps

1. Obtain user input and initialize counter: same as method one.
2. Loop iteration: Same as method one.
3. Checking numbers (mathematically): Inside the loop, for the current number i, create a temporary variable val = i.
   • Use a while loop to continue as long as val > 0.
   • Inside the while loop, get the single digit of val through val % 10.
   • If the ones digit is '1', a boolean flag hasOne = true is set, and the inner while loop is immediately broken out, since it has been determined that the number contains '1'.
   • Otherwise, remove the single digit by val = val / 10 and continue checking the next digit.
4. Skip or output:
   • If hasOne is true, use the continue statement to skip the remainder of the current outer for loop.
   • If hasOne is false (meaning the number does not contain '1'), print the number i and increase count by 1.
5. Close the resource: Close the Scanner object when the program ends.

Sample code

 import java.util.Scanner;

public class NumberFilterMath {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Please enter an integer n:");
        int num = sc.nextInt(); // Get the n entered by the user

        int count = 0; // Initialize the counter of the output numbers System.out.println("The qualified number sequence is as follows:");

        // Loop to find and output n qualified numbers for (int i = 1; count  0) {
                if (val % 10 == 1) { // If the current bit is '1'
                    hasOne = true; // Set the flag break; // Stop checking after finding '1'}
                val = val / 10; // remove current bit, check next bit}

            if (hasOne) {
                continue; // If it contains '1', skip the current number}
            System.out.print(i " "); // Output the number count that meets the conditions; // Add 1 to the counter
        }
        System.out.println("\nA total of " count " numbers were output.");
        sc.close(); // Close Scanner
    }
}

Advantages and Disadvantages

• Advantages: Avoids the overhead of string conversion, which may be more efficient for processing very large numbers or scenarios that require high performance.
• Disadvantages: The logic is slightly more complicated than string conversion and requires an inner while loop to process each bit.

Notes and Summary

1. Loop condition: Both methods use a loop structure such as for (int i = 1; count
2. continue keyword: The continue statement plays an important role here. It allows the program to skip the remaining code of the current loop and directly enter the next iteration when it finds a number that does not meet the conditions, thus avoiding unnecessary processing.
3. Input and Output: Ensure that the program receives user input correctly and outputs the results in a clear format. In practical applications, you can consider storing the output numbers in a list instead of printing them directly for subsequent processing.
4. Resource management: After using the Scanner object, be sure to call sc.close() to close the resource to prevent resource leakage.

In summary, developers can choose a suitable implementation method based on specific needs. If you pursue code simplicity and readability, and do not have extreme performance requirements, the string conversion method is a good choice. If you need to process a large amount of data and have high performance requirements, then mathematical algorithms will be a better solution. Understanding the core logic of these two methods will help you flexibly deal with similar problems in different scenarios.

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