題目如下：

Given a non-empty string containing an out-of-order English representation of digits 0-9, output the digits in ascending order.

Note:

1. Input contains only lowercase English letters.
2. 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.
3. Input length is less than 50,000.

Example 1:

Input: "owoztneoer"

Output: "012"

Example 2:

Input: "fviefuro"

Output: "45"

解題思路：觀察0~9所有的英文表達，可以發現有一些字符只會出現一次，例如z只有在zero中有，w在two中有，整理可以得到如下。

　　     #phase 1
        uniq_dic_1 = {}
        uniq_dic_1[‘z‘] = (‘zero‘,‘0‘)
        uniq_dic_1[
 
‘w‘] = (‘two‘,‘2‘)
        uniq_dic_1[‘u‘] = (‘four‘,‘4‘)
        uniq_dic_1[‘x‘] = (‘six‘,‘6‘)
        uniq_dic_1[‘g‘] = (‘eight‘,‘8‘)

除去這五個單詞後，繼續尋找只有唯一與眾不同字符的單詞，得到如下。

        #phase 2
        uniq_dic_2 = {}
        uniq_dic_2[‘o‘] = (‘one‘,‘1‘)
        uniq_dic_2[‘t‘] = (‘three‘, ‘3‘)
        uniq_dic_2[
 
‘f‘] = (‘five‘, ‘5‘)
        uniq_dic_2[‘s‘] = (‘seven‘, ‘7‘)

除去上面找到的9個，最後就只剩下nine了。

        #phase 3
        uniq_dic_3 = {}
        uniq_dic_3[‘i‘] = (‘nine‘, ‘9‘)

解題的方法，是先去 phase 1 中找出唯一字符在s中出現了幾次，出現了幾次就表示對應的單詞出現了幾次，扣除掉這個單詞其余字符出現的次數；接下來是phase 2和phase 3，即可得到所有單詞出現的次數。

代碼如下：

class Solution(object):
    def calc(self,dic_src,uniq_dic):
        r = ‘‘
        for key in uniq_dic.iterkeys():
            if key in dic_src:
                count = dic_src[key]
                r += uniq_dic[key][1] * count
                for char in uniq_dic[key][0]:
                    dic_src[char] -= count
                    if dic_src[char] == 0:
                        del dic_src[char]
        return r

    def originalDigits(self, s):
        """
        :type s: str
        :rtype: str
        """
        dic_src = {}
        for i in s:
            dic_src[i] = dic_src.setdefault(i, 0) + 1

        #phase 1
        uniq_dic_1 = {}
        uniq_dic_1[‘z‘] = (‘zero‘,‘0‘)
        uniq_dic_1[‘w‘] = (‘two‘,‘2‘)
        uniq_dic_1[‘u‘] = (‘four‘,‘4‘)
        uniq_dic_1[‘x‘] = (‘six‘,‘6‘)
        uniq_dic_1[‘g‘] = (‘eight‘,‘8‘)

        #phase 2
        uniq_dic_2 = {}
        uniq_dic_2[‘o‘] = (‘one‘,‘1‘)
        uniq_dic_2[‘t‘] = (‘three‘, ‘3‘)
        uniq_dic_2[‘f‘] = (‘five‘, ‘5‘)
        uniq_dic_2[‘s‘] = (‘seven‘, ‘7‘)

        #phase 3
        uniq_dic_3 = {}
        uniq_dic_3[‘i‘] = (‘nine‘, ‘9‘)
        res = ‘‘
        res += self.calc(dic_src, uniq_dic_1)
        res += self.calc(dic_src, uniq_dic_2)
        res += self.calc(dic_src, uniq_dic_3)


        return ‘‘.join(sorted(list(res)))

【leetcode】423. Reconstruct Original Digits from English