Reconstruct Original Digits from English

题目来源
给一个字符串，由顺序混乱的英文字母数字组成，然后求将其复原后的情况。
我用了有点取巧的方法，就是按顺序来，因为这些个英文数字的字母可以区分这些数字，代码如下：

class Solution {
public:
    string originalDigits(string s) {
        int n = s.size();
        vector chaCnt(26, 0);
        for (int i=0; i primes{'z', 'x', 's', 'u', 'f', 'h', 'i', 'w', 't', 'n'};
        vector nums{"zero", "six", "seven", "four", "five", "eight", "nine", "two", "three", "one"};
        vector digits{0, 6, 7, 4, 5, 8, 9, 2, 3, 1};
        vector cnt(10, 0);
        for (int i=0; i<10; i++) {
            int tmp = chaCnt[primes[i]-'a'];
            for (int j=0; j

发现方法都差不多，不过答案比我简洁些。
代码如下：

string originalDigits(string s) {
        vector a(10, 0);
        vector alpha(128, 0);
        for (char c : s)
            alpha[c]++;
        a[0] = alpha['z'];
        a[2] = alpha['w'];
        a[4] = alpha['u'];
        a[6] = alpha['x'];
        a[8] = alpha['g'];
        a[3] = alpha['h'] - a[8];
        a[5] = alpha['f'] - a[4];
        a[7] = alpha['v'] - a[5];
        a[1] = alpha['o'] - a[0] - a[2] - a[4];
        a[9] = alpha['i'] - a[5] - a[6] - a[8];
        string ans;
        for (int i = 0; i < 10; i++) {
            if (a[i] > 0)
                ans += string(a[i], '0' + i);
        }
        return ans;
    }