其他题目---数字的英文表达和中文表达

【题目】

　　给定一个32位整数n，写两个函数分别返回n的英文与中文表达字符串。

【基本思路】

　　该类型的题目通常是由小的，简单的场景出发，把复杂的事情拆解成简单的场景。

　　英文表达的实现。英文的表达是以三个数为一组的，所以只要能表达出1~999，然后将数字分解成十亿组，百万组，千组，1～999组，每组都用1~999表达再把组与组之间各自的表达字符串连接起来即可。如何实现1~999的表达？从简单场景出发，依次实现1~19，1~99，1~999即可。

　　需要注意的是负数要加一个前缀，n如果为最小负整数的时候要考虑取绝对值时会溢出的问题。

#python3.5def num1To19(num):    if num < 1 or num > 19:        return ""    names = ["One","Two","Three","Four","Five","Six","Seven","Eight","Nine",\            "Ten","Eleven","Twelve","Thirteen","Fourteen","Fifteen",\            "Sixteen","Seventeen","Eighteen","Nineteen"]    return names[num-1]def num1To99(num):    if num < 1 or num > 99:        return ""    if num >= 1 and num <= 19:        return num1To19(num)    high = num // 10    tyNames = ["Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"]    return tyNames[high-2] + " " + num1To19(num % 10)def num1To999(num):    if num < 1 or num > 999:        return ""    if num >= 1 and num <= 99:        return num1To99(num)    return num1To19(num // 100) + " Hundred " + num1To99(num % 100)def getNumEngExp(num):    if num == 0:        return "Zero"    res = ""    if num < 0:        res = "Negetive, "    if num == -(1 << 31):    #防止取绝对值时溢出        res += "Two Billion, "        num %= -2000000000    num = abs(num)    high = 1000000000    names = ["Billion","Million","Thousand",""]    index = 0    while num != 0:        cur = num // high        num %= high        if cur != 0:            res += num1To999(cur) + " " + names[index]            res += ", " if num != 0 else ""        index += 1        high //= 1000    return res

　　中文表达的实现。与英文表达的处理过程类似，都是由小范围的数向大范围的数扩张。依次实现1~9，1~99，1~999，1~9999，1~99999999。需要注意的细节就是，16的表达为十六，116的表达为一百一十六，1016的表达是一千零一十六，也可以是一千零十六。这说明，对10~19来说，如果百位上有数字，那么就应该念一十几，如果百位没有数字，则应该念十几，所以在设计1~99函数的时候，要多考虑百位是否存在的因素。

def num1To9ByChinese(num):    if num < 1 or num > 9:        return ""    names = ["一","二","三","四","五","六","七","八","九"]    return names[num-1]def num1To99ByChinese(num, hasBai):    if num < 1 or num > 99:        return ""    if num < 10:        return num1To9ByChinese(num)    shi = num // 10    if shi == 1 and not hasBai:        return "十" + num1To9ByChinese(num % 10)    else:        return num1To9ByChinese(shi) + "十" + num1To9ByChinese(num % 10)def num1To999ByChinese(num):    if num < 1 or num > 999:        return ""    if num < 100:        return num1To99ByChinese(num, False)    res = num1To9ByChinese(num // 100) + "百"    rest = num % 100    if rest == 0:        return res    elif rest >= 10:        res += num1To99ByChinese(rest, True)    else:        res += "零" + num1To9ByChinese(rest)    return resdef num1To9999ByChinese(num):    if num < 1 or num > 9999:        return ""    if num < 1000:        return num1To999ByChinese(num)    res = num1To9ByChinese(num // 1000) + "千"    rest = num % 1000    if rest == 0:        return res    elif rest >= 100:        res += num1To999ByChinese(rest)    else:        res += "零" + num1To99ByChinese(rest, False)    return resdef num1To99999999ByChinese(num):    if num < 1 or num > 99999999:        return ""    if num < 10000:        return num1To9999ByChinese(num)    wan = num // 10000    rest = num % 10000    res = num1To9999ByChinese(wan) + "万"    if rest == 0:        return res    elif rest > 1000:        res += num1To9999ByChinese(rest)    else:        res += "零" + num1To999ByChinese(rest)    return resdef getNumChineseExp(num):    if num == 0:        return "零"    res = ""    if num < 0:        res = "负"    num = abs(num)    yi = num // 100000000    rest = num % 100000000    if yi == 0:        return res + num1To99999999ByChinese(rest)    res += num1To99999999ByChinese(yi) + "亿"    if rest == 0:        return res    elif rest > 10000000:        return res + num1To99999999ByChinese(rest)    else:        return res + "零" + num1To99999999ByChinese(rest)