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 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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;
}
};