LeetCode #423 - Reconstruct Original Digits from English

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本文介绍了一种算法，用于解析并还原由英文单词形式表示、顺序混乱的数字串。通过统计字母出现频率并利用特定字母识别对应的数字，该算法能够准确地识别0-9之间的每一个数字，并最终按升序输出。

题目描述：

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"

给定一个字符串，它包含了多个0~9的英文单词，并且字母的顺序已经全部打乱，求出这些数字，并按照从大到小排序。这道题首先需要统计字符串中各个字母的次数，然后根据每个单词的特征判断各个单词的次数，例如字母‘x’是单词“six”特有的，还有字母‘g’是单词“eight”特有的，一些特征不明显的单词需要通过进一步的加减运算求出个数。

class Solution {
public:
    string originalDigits(string s) {
        vector<int> count(26,0);
        for(int i=0;i<s.size();i++) count[s[i]-'a']++;
        vector<int> digit(10,0);
        digit[0]=count['z'-'a'];
        digit[1]=count['o'-'a']-count['z'-'a']-count['u'-'a']-count['w'-'a'];
        digit[2]=count['w'-'a'];
        digit[3]=count['h'-'a']-count['g'-'a'];
        digit[4]=count['u'-'a'];
        digit[5]=count['f'-'a']-count['u'-'a'];
        digit[6]=count['x'-'a'];
        digit[7]=count['s'-'a']-count['x'-'a'];
        digit[8]=count['g'-'a'];
        digit[9]=count['i'-'a']-count['f'-'a']+count['u'-'a']-count['x'-'a']-count['g'-'a'];
        
        string result;
        for(int i=0;i<10;i++)
        {
            while(digit[i]>0)
            {
                result+=(i+'0');
                digit[i]--;
            }
        }
        return result;
    }
};