Given a non-empty string containing an out-of-order English representation of digits 0-9, output the digits in ascending order.
Note:
- Input contains only lowercase English letters.
- Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.
- Input length is less than 50,000.
Example 1:
Input: "owoztneoer" Output: "012"
Example 2:
Input: "fviefuro" Output: "45"
class Solution(object):
def originalDigits(self, s):
"""
:type s: str
:rtype: str
six, zero, two, eight, four中分别包含唯一字母x, z, w, g, u, 根据z,w,x,g,u的个数就可以知道0,2,6,8,4的个数。对于剩下的one,three,five,seven，one可以由字符o的个数减去在0,2,4,6,8中出现的o的个数。
"""
char_cnt = [0]*26
for c in s:
char_cnt[ord(c)-ord('a')] += 1
mapp_digits = {"zero": 0, "one": 1, "two": 2, "three": 3, "four": 4, "five": 5, "six": 6, "seven": 7, "eight": 8, "nine": 9}
digits_c = {"zero": "z", "two": "w", "four": "u", "six": "x", "eight": "g"}
digits_c2 = {"one": "o", "three": "h", "five": "f", "seven": "s"}
digits_c3 = {"nine": "i"}
out = []
for digits in (digits_c, digits_c2, digits_c3):
for d in digits:
cnt = char_cnt[ord(digits[d])-ord('a')]
if cnt > 0:
out += [mapp_digits[d]] * cnt
for c in d:
char_cnt[ord(c)-ord('a')] -= cnt
out.sort()
return "".join(str(c) for c in out)