Reconstruct Original Digits from English

题目：

Given a non-empty string containing an out-of-order English representation of digits0-9, output the digits in ascending order.

Note:

1. Input contains only lowercase English letters.
2. 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.
3. Input length is less than 50,000.

Example 1:

Input: "owoztneoer"Output: "012"

Example 2:

Input: "fviefuro"Output: "45"

代码：

class Solution {public:    string originalDigits(string s) {        int casetable[30];        memset(casetable,0,sizeof(casetable));        for(int i=0; i<s.length(); i++)        {            casetable[s[i]-'a']++;        }        string ans="";        int a[11];        a[0]=casetable['z'-'a'];        a[2]=casetable['w'-'a'];        a[4]=casetable['u'-'a'];        a[6]=casetable['x'-'a'];        a[8]=casetable['g'-'a'];        a[1]=casetable['o'-'a']-a[2]-a[4]-a[0];        a[3]=casetable['h'-'a']-a[8];        a[5]=casetable['f'-'a']-a[4];        a[7]=casetable['s'-'a']-a[6];        a[9]=casetable['i'-'a']-a[5]-a[6]-a[8];        for(int i=0; i<10; i++)        {            ans+=string(a[i],'0'+i);        }        return ans;    }};

分析： 找到所有数字中独有的字母，在字符串中该字符的数量就是该数字的数量。