1. 题目
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)
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reconstruct-original-digits-from-english
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2. 解题
- 先找出只唯一拥有某字符的单词
- Zero，tWo，foUr，siX，eiGht
- 然后上面找完了，再找剩下的唯一拥有字符的
- One，Three，Five，Seven
- 最后留下 nine
class Solution {
public:string originalDigits(string s) {int count[26] = {0};for(auto& ch : s)count[ch-'a']++;int num[10] = {0};num[0] = count['z'-'a'];count['z'-'a'] -= num[0];count['e'-'a'] -= num[0];count['r'-'a'] -= num[0];count['o'-'a'] -= num[0];num[2] = count['w'-'a'];count['t'-'a'] -= num[2];count['w'-'a'] -= num[2];count['o'-'a'] -= num[2];num[4] = count['u'-'a'];count['f'-'a'] -= num[4];count['o'-'a'] -= num[4];count['u'-'a'] -= num[4];count['r'-'a'] -= num[4];num[6] = count['x'-'a'];count['s'-'a'] -= num[6];count['i'-'a'] -= num[6];count['x'-'a'] -= num[6];num[8] = count['g'-'a'];count['e'-'a'] -= num[8];count['i'-'a'] -= num[8];count['g'-'a'] -= num[8];count['h'-'a'] -= num[8];count['t'-'a'] -= num[8];num[1] = count['o'-'a'];count['o'-'a'] -= num[1];count['n'-'a'] -= num[1];count['e'-'a'] -= num[1];num[3] = count['t'-'a'];count['t'-'a'] -= num[3];count['h'-'a'] -= num[3];count['r'-'a'] -= num[3];count['e'-'a'] -= 2*num[3];num[5] = count['f'-'a'];count['f'-'a'] -= num[5];count['i'-'a'] -= num[5];count['v'-'a'] -= num[5];count['e'-'a'] -= num[5];num[7] = count['s'-'a'];count['s'-'a'] -= num[7];count['e'-'a'] -= 2*num[7];count['v'-'a'] -= num[7];count['n'-'a'] -= num[7];num[9] = count['i'-'a'];string ans;for(int i = 0; i < 10; ++i){while(num[i]--)ans += to_string(i);}return ans;}
};
