423. Reconstruct Original Digits from English

https://leetcode.com/problems/reconstruct-original-digits-from-english/description/

Description

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"

Solution

Solution-1

使用动态规划方式求解。但是在输入非常大时，迭代的层数会非常多，空间复杂度会非常大。

class Solution(object):
    def recursive_f(self, letters):
        if len(letters)==0: return True, ''
        for s_digit, counter in self.record:
            for letter in counter:
                if counter[letter]>letters[letter]:
                    break
            else:
                r,v = self.recursive_f(letters-counter)
                if r:
                    return True, s_digit+v
        else:
            return False, None
    def originalDigits(self, s):
        """
        :type s: str
        :rtype: str
        """
        from collections import Counter
        self.record = [(n,Counter(v)) for n,v in zip(list('0123456789'),
                        ['zero','one','two','three','four','five',
                         'six','seven','eight','nine'])]
        letters = Counter(s)
        r, v = self.recursive_f(letters)
        return v

Solution-2

根据题目特性，比如：在 0~9 对应的字母中，只有 0（zero）有 z, 因此可以先求出 0 的个数，以此类推可以大大加快求解效率。

class Solution(object):
    def originalDigits(self, s):
        """
        :type s: str
        :rtype: str
        """
        from collections import Counter
        record = [0]*10
        letters = Counter(s)
        items = [(0, 'z', 'zero'), (6, 'x', 'six'), (2, 'w', 'two'),
                 (8, 'g', 'eight'), (3, 'h', 'three'), (4, 'u', 'four'),
                 (5, 'f', 'five'), (1, 'o', 'one'), (9, 'i', 'nine'),
                 (7, 's', 'seven')]
        for n, flag, word in items:
            if letters[flag]:
                record[n] += letters[flag]
                v = letters[flag]
                for letter in word: letters[letter] -= v
        return ''.join([str(i)*v for i,v in enumerate(record)])