网易公司刁难的面试题,留下你的见解

哈哈  火星兄很无奈的说

首先，英文单词是3位3位来的。分别是million, thousand和啥都没有的后缀。

然后问题就变成解析一个三位数的问题了，譬如，123翻译成英文。

这就简单了，%100, %10, %1，得出的值，对应到3个数组里面去。。。


面试题嘛，就这样说行了，不用给代码了吧？这很简单阿。。。

搞个计数器判断下就可以了，和汉字按首字母排序差不多的问题

101/1001这样的也是要考虑的吧

这道题很简单，如果这也不会的话，转行做QA吧，真的。。。

是关于我给的那个算法

的问题吗？

这种情况已经考虑进来的阿。

1001-〉分成2档，1 -> thousand, 001 -> 啥都没有

1/100 = 0, ["","one hundred and ", "two hundred and ", "three hundred and "..."nine hundred and "]

(1-0*100)/10 = 0, ["", "ninty" ...]

(1-0*100-1*10)/1 = 1 ["", "nine"...]

那个%100,%10,%1是指那个经典的算法，学校里面求每位是几那个。

想问网易是否诚心招人啊？

越来越不理解了....

如果那么简单,请把代码帖出来吧

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;    
    }    
   
}