Given a non-empty string containing an out-of-order English representation of digits 0-9, output the digits in ascending order.
Note:
- Input contains only lowercase English letters.
- 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.
- Input length is less than 50,000.
Example 1:
Input: "owoztneoer" Output: "012"
Example 2:
Input: "fviefuro" Output: "45"
这道题给了我们一串英文字符串，是由表示数字的英文单词组成的，不过字符顺序是打乱的，让我们重建出数字。那么这道题的思路是先要统计出各个字符出现的次数，然后算出每个单词出现的次数，然后就可以重建了。由于题目中限定了输入的字符串一定是有效的，那么不会出现无法成功重建的情况，这里需要用个trick。我们仔细观察这些表示数字的单词"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"，我们可以发现有些的单词的字符是独一无二的，比如z，只出现在zero中，还有w，u，x，g这四个单词，分别只出现在two，four，six，eight中，那么这五个数字的个数就可以被确定了，由于含有o的单词有zero，two，four，one，其中前三个都被确定了，那么one的个数也就知道了；由于含有h的单词有eight，three，其中eight个数已知，那么three的个数就知道了；由于含有f的单词有four，five，其中four个数已知，那么five的个数就知道了；由于含有s的单词有six，seven，其中six个数已知，那么seven的个数就知道了；由于含有i的单词有six，eight，five，nine，其中前三个都被确定了，那么nine的个数就知道了，知道了这些问题就变的容易多了，我们按这个顺序"zero", "two", "four", "six", "eight", "one", "three", "five", "seven", "nine"就能找出所有的个数了，参见代码如下：
解法一：
class Solution {
public:
string originalDigits(string s) {
string res = "";
vector<string> words{"zero", "two", "four", "six", "eight", "one", "three", "five", "seven", "nine"};
vector<int> nums{, , , , , , , , , }, counts(, );
vector<char> chars{'z', 'w', 'u', 'x', 'g', 'o', 'h', 'f', 's', 'i'};
for (char c : s) ++counts[c - 'a'];
for (int i = ; i < ; ++i) {
int cnt = counts[chars[i] - 'a'];
for (int j = ; j < words[i].size(); ++j) {
counts[words[i][j] - 'a'] -= cnt;
}
while (cnt--) res += (nums[i] + '');
}
sort(res.begin(), res.end());
return res;
}
};
另外我们也可以用更加简洁易懂的方法来快速的找出各个数字的个数，参见代码如下：
解法二：
class Solution {
public:
string originalDigits(string s) {
string res = "";
vector<int> counts(, ), nums(, );
for (char c : s) ++counts[c];
nums[] = counts['z'];
nums[] = counts['w'];
nums[] = counts['u'];
nums[] = counts['x'];
nums[] = counts['g'];
nums[] = counts['o'] - nums[] - nums[] - nums[];
nums[] = counts['h'] - nums[];
nums[] = counts['f'] - nums[];
nums[] = counts['s'] - nums[];
nums[] = counts['i'] - nums[] - nums[] - nums[];
for (int i = ; i < nums.size(); ++i) {
for (int j = ; j < nums[i]; ++j) {
res += (i + '');
}
}
return res;
}
};
参考资料：
https://discuss.leetcode.com/topic/64150/straightforward-c-accepted-solution
https://discuss.leetcode.com/topic/63382/share-my-simple-and-easy-o-n-solution