LeetCode 423. Reconstruct Original Digits from English-优快云博客

本文介绍了一种从乱序的英文数字表述中提取并按升序排列数字的算法。该算法通过分析特定字符出现的次数来确定每个数字出现的频次，并最终输出排序后的数字串。文章详细解释了算法的具体实现步骤。

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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"

class Solution {
public:
    string originalDigits(string s) {
        vector<int> letters(26,0);
        for (auto c: s) {
            ++letters[c - 'a']; 
        }
        vector<string> words = {"zero", "two", "four", "six", "eight", "one", "three", "five", "seven", "nine"};
        vector<int> order = {0, 2, 4, 6, 8, 1, 3, 5, 7, 9};
        vector<char> dis_char = {'z', 'w', 'u', 'x', 'g', 'o', 'r', 'f', 'v', 'i'};
        vector<int> counts(10, 0);
        for (int i = 0;i < 10; ++i) {
            int num = letters[dis_char[i] - 'a'];
            counts[order[i]] = num;
            for (auto c: words[i]) {
                letters[c - 'a'] -= num;
            }
        }
        string result;
        for (int i = 0; i < 10; ++i) {
            result += string(counts[i], char(i +'0'));
        }
        return result;
    }
};