项目Euler #17 (java)

Question

我在Project Euler中从事Problem 17的工作。使用下面的代码，我得到的答案是21088，这是错误的。我不知道问题出在哪里。有人知道我哪里错了吗？谢谢。

public class Problem17 {

    public static void main(String[] args) {
        System.out.println(countLetters(1000));
    }

    public static long countLetters(int n) {
        long sum = 0;

        for (int i=1; i<=n; i++){
            sum += getLetters(i);
        }

        return sum;
    }

    public static int getLetters(int n) {
        int letters = 0;

        switch (n) {
            case 0: letters = 0; break;
            case 1: letters = "One".length(); break;
            case 2: letters = "Two".length(); break;
            case 3: letters = "Three".length(); break;
            case 4: letters = "Four".length(); break;
            case 5: letters = "Five".length(); break;
            case 6: letters = "Six".length(); break;
            case 7: letters = "Seven".length(); break;
            case 8: letters = "Eight".length(); break;
            case 9: letters = "Nine".length(); break;
            case 10: letters = "Ten".length(); break;
            case 11: letters = "Eleven".length(); break;
            case 12: letters = "Twelve".length(); break;
            case 13: letters = "Thirteen".length(); break;
            case 14: letters = "Fourteen".length(); break;
            case 15: letters = "Fifteen".length(); break;
            case 16: letters = "Sixteen".length(); break;
            case 17: letters = "Seventeen".length(); break;
            case 18: letters = "Eighteen".length(); break;
            case 19: letters = "Nineteen".length(); break;
            case 20: letters = "Twenty".length(); break;
            case 30: letters = "Thirty".length(); break;
            case 40: letters = "Forty".length(); break;
            case 50: letters = "Fifty".length(); break;
            case 60: letters = "Sixty".length(); break;
            case 70: letters = "Seventy".length(); break;
            case 80: letters = "Eighty".length(); break;
            case 90: letters = "Ninety".length(); break;
            case 100: letters = "OneHundred".length(); break;
            case 200: letters = "TwoHundred".length(); break;
            case 300: letters = "ThreeHundred".length(); break;
            case 400: letters = "FourHundred".length(); break;
            case 500: letters = "FiveHundred".length(); break;
            case 600: letters = "SixHundred".length(); break;
            case 700: letters = "SevenHundred".length(); break;
            case 800: letters = "EightHundred".length(); break;
            case 900: letters = "NineHundred".length(); break;
            case 1000: letters = "OneThousand".length(); break;
        }

        if (letters == 0){
            int length = (int)(Math.log10(n)+1); //how many digits does the number have? 
            int splitNumber;
            for (int i=1;i<=length;i++){
                if (i == 3) letters += 3; //add "and" to the total letters if necessary
                splitNumber = (int) (n % Math.pow(10, i)); 
                letters += getLetters(splitNumber);
                n -= splitNumber;
            }
        }

        return letters;
    }
}

编辑:刚刚注意到它现在计算的例子是“111”D‘’OH！今晚晚些时候才会有时间调整，到时候会更新。

Answer 1

您的问题是，如果您的案例列表中没有匹配，那么您总是将数字拆分为单独的数字，从较低的数字开始。所以，对于114，你得到4，剩下110，然后你得到10，剩下100。如果你以相反的方式来做，那么对于114，你去掉了最高的部分(100)，那么你只剩下14个，这将与你的交换机一起工作。

顺便说一句，这个问题有很大的对称性，可能只需要几分钟的时间来处理纸和笔。