423 Reconstruct Original Digits from English 从英文中重建数字

给定一个非空字符串，其中包含字母顺序打乱的英文单词表示的数字0-9。按升序输出原始的数字。
注意:
    输入只包含小写英文字母。
    输入保证合法并可以转换为原始的数字，这意味着像 "abc" 或 "zerone" 的输入是不允许的。
    输入字符串的长度小于 50,000。
示例 1:
输入: "owoztneoer"
输出: "012" (zeroonetwo)
示例 2:
输入: "fviefuro"
输出: "45" (fourfive)
详见：https://leetcode.com/problems/reconstruct-original-digits-from-english/description/

C++：

方法一：

class Solution {
public:
    string originalDigits(string s) {
        string res = "";
        vector<int> counts(128, 0), nums(10, 0);
        for (char c : s)
        {
            ++counts[c];
        }
        nums[0] = counts['z'];
        nums[2] = counts['w'];
        nums[4] = counts['u'];
        nums[6] = counts['x'];
        nums[8] = counts['g'];
        nums[1] = counts['o'] - nums[0] - nums[2] - nums[4];
        nums[3] = counts['h'] - nums[8];
        nums[5] = counts['f'] - nums[4];
        nums[7] = counts['s'] - nums[6];
        nums[9] = counts['i'] - nums[6] - nums[8] - nums[5];
        for (int i = 0; i < nums.size(); ++i)
        {
            for (int j = 0; j < nums[i]; ++j)
            {
                res += (i + '0');
            }
        }
        return res;
    }
};

方法二：

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

参考：https://www.cnblogs.com/grandyang/p/5996239.html