【leetcode】423. Reconstruct Original Digits from English

题目如下：

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"

解题思路：观察0~9所有的英文表达，可以发现有一些字符只会出现一次，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')

除去这五个单词后，继续寻找只有唯一与众不同字符的单词，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个，最后就只剩下nine了。

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

解题的方法，是先去 phase 1 中找出唯一字符在s中出现了几次，出现了几次就表示对应的单词出现了几次，扣除掉这个单词其余字符出现的次数；接下来是phase 2和phase 3，即可得到所有单词出现的次数。

代码如下：

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)))