如何找到最小的 integer 优于 integer n？ [关闭]

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I have an assessment. I tried many times to solve this. But I couldn't. Please someone help me. Here is the assessment:

The next function should return the smallest integer which is superior to n and whose digits are all different from all of n' s digits.

For example next(654321) should return 700000.

if no such integer exists then the function must return -1.

write the body of the next(n) function

Note: n is a strictly positive integer lower than 2^31

Answer 1

Try this simple Console App. I wrote in C# (you specified both C# and Java in question tags).

Note: updated function further on

Version 1: using Lists and Enumberable operations

internal class Program
{
    private static readonly double maxValue = Math.Pow(2, 31);  // Max allowed value for function

    private static void Main(string[] args)
    {
        while (true)
        {
            Console.WriteLine("Please specify start value.");
            int startValue = Convert.ToInt32(Console.ReadLine());
            int result = YourAssesmentFunction(startValue);
            Console.WriteLine($"Your function result is {result}.");
        }
    }

    /// <summary>
    /// The next function should return the smallest integer which is superior to n and whose digits are all different from all of n' s digits.
    /// For example next(654321) should return 700000.
    /// if no such integer exists then the function must return -1.
    /// write the body of the next(n) function
    /// Note: n is a strictly positive integer lower than 2^31
    /// </summary>
    private static int YourAssesmentFunction(int startNumber)
    {
        List<char> startChars = new(); // List of char (numbers) contained in given number
        List<char> newChars = new(); // List of char (numbers) contained in next number

        IEnumerable<char>? inCommon = null; // IEnumerable of char used to find common chars

        int nextNumber = -1; // Initialize result with default value -1

        startNumber.ToString().ToList().ForEach(x => startChars.Add(x)); // Use Linq to populate list of digits in given number

        while (inCommon == null || inCommon.Any()) // The first execution has inCommon == null as entry condition
        {
            if (startNumber >= maxValue || startNumber < 0) // If maxValue has been reached or given number is less than 0, exit while loop
                break;

            nextNumber = startNumber + 1;
            newChars.Clear(); // Clear the list from previous values

            nextNumber.ToString().ToList().ForEach(x => newChars.Add(x)); // Use Linq to populate list of digits in next number

            inCommon = startChars.Intersect(newChars).ToList(); // Use Linq to find common digits

            startNumber++; // Increment startNumber for the next while loop execution
        }
        return nextNumber;
    }
}

The code you need is in YourAssesmentFunction() . Please note that I'm a self-made student programmer and there might be much better ways to write this function.

EDIT: After Klaus Gutter's comment, I tried to find a simpler, faster and more elegant way of writing the same function. Here is it:

Version 2: using loop through digits in trying number

    private static int YourAssesmentFunctionOptimized(int startNumber)
    {
        int tryNext = startNumber + 1; // Next number to try

        // String representation of given number
        string start = startNumber.ToString();

        // Flags for while-loop managing
        bool numberFound = false;
        bool skipToNext;

        while (!numberFound)
        {
            if (tryNext > maxValue | startNumber < 0) // If number to try is greater than maxValue (2^31) or start number is less than 0, exit and return 0
                return -1;
            skipToNext = false;
            foreach (char c in tryNext.ToString()) // Loop through chars (= digits) in string representation of trying number
            {
                if (start.Contains(c))
                {
                    tryNext++; // If a digit of the trying number is found in given number, the trying number is not the result
                    skipToNext = true;
                    break; // Exit foreach loop and skip to next number
                }
            }
            if (!skipToNext)
                numberFound = true; // If skip to next = true, numberFound remains false and while loop keep executing
        }
        return tryNext;
    }

This function is much faster because it doesn't need to create List s or evaluating IEnumerable operations.