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
     
       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
     
       words{"zero", "two", "four", "six", "eight", "one", "three", "five", "seven", "nine"};        vector
      
        nums{0, 2, 4, 6, 8, 1, 3, 5, 7, 9}, counts(26, 0);        vector
       
         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