Question

public class NumberToWords2 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int n = 30001;
        numberToWords(n);
    }

    public static String numberToWords(int n){

        String temp = Integer.toString(n);
        int[] myArr = new int[temp.length()];
        for (int i = 0; i < temp.length(); i++)
        {
            myArr[i] = temp.charAt(i) - '0';
        }

        if(myArr.length == 1){
            System.out.println(oneDigits(n));
        }
        else if(myArr.length == 2){
            System.out.println(twoDigits(n));
        }
        else if(myArr.length == 3){
            System.out.println(threeDigits(n));
        }
        else if(myArr.length == 4){
            System.out.println(fourDigits(n));
        }
        else if(myArr.length == 5){
            System.out.println(fiveDigits(n));
        }

        return "Invalid Input";

    }

    ///// Methods to return the equivalent English words. Logic and "And" "Thousands" etc /////

    private static String fiveDigits(int n) {
        // TODO Auto-generated method stub

        //Check for 20000, 30000, 40000 etc
        if(n%1000 == 0){
            return twoDigits(n/1000) + " Thousand";
        }

        //Numbers starting with 1
        if(n/10000 == 1){
            //Handle numbers like 10001, 10002, 70024, 80099 etc 
            if(n%1000 < 100){
                return ones(n/1000) + " Thousand And " + twoDigits(n%1000);
            }
            else{
                return ones(n/1000) + " Thousand " + threeDigits(n%1000);
            }
        }
        else{
            if(n%1000 < 100){
                return twoDigits(n/1000)  + " Thousand And " + twoDigits(n%1000);
            }else{
                return twoDigits(n/1000) + " Thousand " + threeDigits(n%1000);
            }           
        }
    }

    private static String fourDigits(int n) {
        // TODO Auto-generated method stub

        //Check for 2000, 3000, 4000 etc
        if(n%1000 == 0){
            return ones(n/1000) + " Thousand";
        }
        //Handle numbers like 1001, 1002, 7024, 8099 etc 
        else if(n%1000 < 100){
            return ones(n/1000) + " Thousand And " + twoDigits(n%1000);
        }
        //Normal Case
        else{
            return ones(n/1000) + " Thousand " + threeDigits(n%1000);
        }

    }

    private static String threeDigits(int n) {
        // TODO Auto-generated method stub

        //Check for 200, 300, 400 etc
        if(n%100 == 0){
            return ones(n/100) + " Hundred";
        }
        //Normal Case
        else{
            return ones(n/100) + " Hundred And " + twoDigits(n%100);
        }
    }

    private static String twoDigits(int n) {
        // TODO Auto-generated method stub

        //Check for 11, 12, 13, 14 etc OR Handle Single digit so can reuse code
        if(n/10 == 1 || n/10 == 0)
            return ones(n);
        //Check for 20, 30, 40 etc. Cannot print zero at the back
        else if(n%10 == 0){
                return tens(n/10);
        }
        //Normal Case
        else{   
            return tens(n/10) + " " + ones(n%10);   
        }
    }

    private static String oneDigits(int n) {
        // TODO Auto-generated method stub
        return ones(n);
    }

    ///// Return number words only /////

    private static String ones(int num){

        Map<Integer, String> h = new HashMap<Integer, String>();

        h.put(0 , "Zero");
        h.put(1 , "One");
        h.put(2 , "Two");
        h.put(3 , "Three");
        h.put(4 , "Four");
        h.put(5 , "Five");
        h.put(6 , "Six");
        h.put(7 , "Seven");
        h.put(8 , "Eight");
        h.put(9 , "Nine");
        h.put(10, "Ten");
        h.put(11 , "Eleven");
        h.put(12 , "Twelve");
        h.put(13 , "Thirteen");
        h.put(14 , "Fourteen");
        h.put(15 , "Fifteen");
        h.put(16 , "Sixteen");
        h.put(17 , "Seventeen");
        h.put(18 , "Eighteen");
        h.put(19 , "Nineteen");

        return h.get(num);

    }

    private static String tens(int num){

        Map<Integer, String> h = new HashMap<Integer, String>();

        h.put(2 , "Twenty");
        h.put(3 , "Thirty");
        h.put(4 , "Fourty");
        h.put(5 , "Fifty");
        h.put(6 , "Sixty");
        h.put(7 , "Seventy");
        h.put(8 , "Eighty");
        h.put(9 , "Ninety");

        return h.get(num);

    }
}

我试图学习使用更简单，更优雅的方法来改进我的代码编写的方法来执行条件if else语句。有没有什么方法可以改进这个代码，使其像专业人士一样更短，更易读？

Answer 1

我希望它可以帮到你

    int numOfDigits = n/1000;
    String result = String.valueOf(numOfDigits == 1? ones(numOfDigits) : twoDigits(numOfDigits));

    String resultString = result + " Thousand " + twoDigits(n%1000);

    if(n%1000 < 100){
        resultString = result + " Thousand And " + threeDigits(n%1000);
    }

    return resultString;

Answer 2

不知道你是否认为这更好，但这里有一个建议:(完全未经测试，但你明白了）

String baseStr;
if(n/10000 == 1) {
    //Handle numbers like 10001, 10002, 70024, 80099 etc 
    baseStr = ones(n/1000);
} else {
    baseStr = twoDigits(n/1000);
}
return baseStr + " Thousand " + ((n%1000 < 100) ? "And " twoDigits(n%1000) : + threeDigits(n%1000));

Answer 3

我猜你可以像下面那样做空。

if(n/10000 == 1 && n%1000 < 100) {
    return ones(n/1000) + " Thousand And " + twoDigits(n%1000);
}else if(n%1000 < 100) {
    return twoDigits(n/1000)  + " Thousand And " + twoDigits(n%1000);
}else{
    return twoDigits(n/1000) + " Thousand " + threeDigits(n%1000);           
}

Answer 4

您可以通过重构逻辑并逐个构建结果来获得更简单的代码：

    String result = "";
    if (n/10000 == 1) {
        //Handle numbers like 10001, 10002, 70024, 80099 etc 
        result += ones(n/1000) + " Thousand";
    } else {
        result += twoDigits(n/1000)  + " Thousand";
    }
    if (n%1000 < 100) {
        result += " And " + twoDigits(n%1000);
    } else {
        result += " " + threeDigits(n%1000);
    }
    return result;

Answer 5

您可以使用两个三元运算符使其成为单个表达式。为了便于阅读，您还可以将n / 1000和n % 1000的值存储在局部变量中，以避免重复。

int q = n / 1000;
int r = n % 1000;
return (q / 10 == 1 ? ones(q) : twoDigits(q)) + " Thousand "
        + (r < 100 ? "And " + twoDigits(r) : threeDigits(r));

Answer 6

您可以保存计算，提供有意义的名称并正确缩进代码，如下所示：

    int nOverThousand = n / 1000;
    int nPercThousand = n % 1000;

    int twoDigitsTerm = twoDigits(nPercThousand);
    int threeDigitsTerm = threeDigits(nPercThousand);

    if(nOverThousand / 10 == 1){

        //Handle numbers like 10001, 10002, 70024, 80099 etc 
        int term1 = ones(nOverThousand);
        return (nPercThousand < 100) ?
                (term1 + " Thousand And " + twoDigitsTerm) :
                (term1 + " Thousand " + threeDigitsTerm);
    }
    else{
        int term1 = twoDigits(nOverThousand);
        return (nPercThousand < 100) ?
                (term1 + " Thousand And " + twoDigitsTerm) :
                (term1 + " Thousand " + threeDigitsTerm);
    }

请注意，

保留可读性。有意义的名字使它更具可读性。
节省计算费用。
代码看起来比较紧凑。

编写更短的if else代码

6 个答案: