Convert a non-negative integer num to its English words representation.
Example 1:
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Input: num = 123
Output: "One Hundred Twenty Three"
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Example 2:
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Input: num = 12345
Output: "Twelve Thousand Three Hundred Forty Five"
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Example 3:
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Input: num = 1234567
Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"
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Constraints:
解题思路：
根据数字长度分成不同的翻译位:
亿，百万，千，百，Billion, Million, Thousand, Hundred
90-20的整10读法: Ninety, Eighty, Seventy, Sixty, Fifty, Fourty, Thirty, Twenty
19-11的读法：Nienteen, Eighteen, Seventeen, Sixteen, Fifteen, Fourteen, Thirteen, Twelve, Eleven
10-0的读法：Ten, Nine, Eight, Seven, Six, Five, Four, Three, Two, One, Zero
难点：把数字根据位数拆开。
**两种解法：**递归和迭代。
递归解法：
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class Solution(object):
def numberToWords(self, num):
"""
:type num: int
:rtype: str
"""
from0to19=['','One','Two','Three','Four','Five','Six','Seven','Eight','Nine','Ten','Eleven','Twelve','Thirteen','Fourteen','Fifteen','Sixteen','Seventeen','Eighteen','Nineteen']
from20to90=['','','Twenty','Thirty','Forty','Fifty','Sixty','Seventy','Eighty','Ninety']
def recursive(num):
if num>=1000000000: # larger than 1 billion
ans='{}Billion'.format(from0to19[num//1000000000])+recursive(num%1000000000)
return ans
elif num>=1000000:
ans='{}Million'.format(recursive(num//1000000))+recursive(num%1000000)
return ans
elif num>=1000:
ans='{}Thousand'.format(recursive(num//1000))+recursive(num%1000)
return ans
elif num>=100:
ans='{}Hundred'.format(from0to19[num//100])+recursive(num%100)
return ans
elif num>=20:
ans='{}'.format(from20to90[num//10])+recursive(num%10)
return ans
else:
return '{}'.format(from0to19[num])
def addBlank(s):
_s=''
for _ in s:
if _.isupper():
_s+=' {}'.format(_)
else:
_s+=_
return _s
if num==0:
return 'Zero'
else:
ans=recursive(num)
return addBlank(ans)[1:]
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Runtime: 24 ms, faster than 54.38% of Python online submissions for Integer to English Words.
Memory Usage: 13.6 MB, less than 13.33% of Python online submissions for Integer to English Words.
踩坑：
- 空格问题
- 数字区间分割
- Billion位忘记转英文
- 英文拼写错误，eg: forteen