Reconstruct Original Digits from English

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首先，统计字符串s中每个字母出现的次数，然后为nums中的每个数字记录数字计数。例如，对于数字0，dcounts[0] = lcounts[distinct[0] - ‘a’]，这表示在字符串中'z'出现了多少次，即“zero”的数量。接着，对于“zero”中的每个字母，减少lcounts的计数，因为这些字母将用于重构数字。

First, calculate how many times each letter appeared in the string $s$ , and then for each number in the $nums$ record digit counts. For example, for number 0, dcounts[0] = lcounts[distinct[0] - ‘a’], which represents how many times ‘z’ appeared in the string; in other words, how many “zero” in the string. Next for every letter in “zero”, reduce the times from $lcounts$ , because these letters will be used in reconstructing the digits.

class Solution {
public:
    string originalDigits(string s) {
        vector<string> words = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
        vector<int> nums = {0, 2, 4, 6, 8, 1, 3, 5, 7, 9};
        vector<int> distinct = {'z', 'o', 'w', 'h', 'u', 'f', 'x', 's', 'g', 'i'};
        vector<int> lcounts(26, 0);
        vector<int> dcounts(10, 0);
        string res = "";
        for(auto c : s) lcounts[c - 'a']++;
        for(auto num : nums){
            dcounts[num] = lcounts[distinct[num] - 'a'];
            for(char c : words[num])
                lcounts[c - 'a'] -= dcounts[num];
        }
        for(int i = 0; i < 10; i++)
            for(int j = 0; j < dcounts[i]; j++)
                res += to_string(i);
        return res;
    }
};