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网易公司刁难的面试题,留下你的见解

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作者 正文
   发表时间:2009-09-10  
确实不是刁难啊 英文是以三作为一个循环 不停的以3位作为一个整体进行匹配 比如 123 就是 one hundred twenty three 那么假如之前还有321 就直接在刚才的123前 + 上 thousand and 然后在匹配啊 3位的匹配很好写的

真的不知道LZ为什么说刁难 还是我理解有误
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   发表时间:2009-09-10  
dingjun123 写道
面试题嘛,转个别人写的
package name.zrl;    
   
public class ListTest {    
    static String[] to_19 = { "zero", "one", "two", "three", "four", "five",    
            "six", "seven", "eight", "nine", "ten", "eleven", "twelve",    
            "thirteen", "fourteen", "fifteen", "sixteen", "seventeen",    
            "eighteen", "nineteen" };    
    static String[] tens = { "twenty", "thirty", "forty", "fifty", "sixty",    
            "seventy", "eighty", "ninety" };    
    static String[] denom = { "", "thousand ", "million", "billion",    
            "trillion", "quadrillion", "quintillion", "sextillion",    
            "septillion", "octillion", "nonillion", "decillion", "undecillion",    
            "duodecillion", "tredecillion", "quattuordecillion",    
            "sexdecillion", "septendecillion", "octodecillion",    
            "novemdecillion", "vigintillion" };    
   
    public static void main(String[] argv) throws Exception {    
        long tstValue = 1234567890987654321L;    
        ListTest itoe = new ListTest();    
        System.out.println(itoe.english_number(tstValue));    
    }    
   
    private String convert_nn(int val) {    
        if (val < 20)    
            return to_19[val];    
   
        int flag = val / 10 - 2;    
        if (val % 10 != 0)    
            return tens[flag] + "-" + to_19[val % 10];    
        else   
            return tens[flag];    
    }    
   
    private String convert_nnn(int val) {    
        String word = "";    
        int rem = val / 100;    
        int mod = val % 100;    
        if (rem > 0) {    
            word = to_19[rem] + " hundred ";    
   
        }    
        if (mod > 0) {    
            word = word + convert_nn(mod);    
        }    
        return word;    
    }    
   
    public String english_number(long val) {    
   
        if (val < 100) {    
            System.out.println((int) val);    
            return convert_nn((int) val);    
        }    
        if (val < 1000) {    
            return convert_nnn((int) val);    
        }    
   
        for (int v = 0; v < denom.length; v++) {    
   
            int didx = v - 1;    
            long dval = new Double(Math.pow(1000, v)).longValue();    
   
            if (dval > val) {    
                long mod = new Double(Math.pow(1000, didx)).longValue();    
                int l = (int) (val / mod);    
                long r = (long) (val - (l * mod));    
                String ret = convert_nnn(l) + " " + denom[didx];    
                if (r > 0) {    
                    ret = ret + ", " + english_number(r);    
                }    
   
                return ret;    
            }    
   
        }    
        return null;    
    }    
   
}    

blue兄,那么晚还回帖真敬业啊。
这个版本有些问题,缺少and,没考虑101这种情况吧。

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