[LeetCode] Reconstruct Original Digits from English

Reconstruct Original Digits from English

Given a non-empty string containing an out-of-order English representation of digits 0-9, 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"

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快两年没更新博客了，今天看了一下Leetcode，都500多道题了，就随便交了一道。

开始考虑到用DFS，但是考虑到有些Digit可以由一个字母唯一确认，然后推算了一下，发现所有Digit都可以确定，就直接算答案了。好久没刷过题了，代码质量直线下降啊。

class Solution {
public:
    void update(int* count, string s, int n) {
        for (auto c : s) {
            count[c - 'a'] -= n;
        }
    }

    string originalDigits(string s) {
        vector<string> dict = {"zero", "one", "two", "three", "fore", "five", "six", "seven", "eight", "nine"};
        int digitOrder[10] = {0, 8, 6, 2, 3, 7, 5, 4, 1, 9};
        char keyOrder[10] = {'z', 'g', 'x', 'w', 'h', 's', 'v', 'f', 'o', 'i'};
        int count[26] = {0};
        for (auto c : s) {
            ++count[c - 'a'];
        }
        int res[10] = {0};
        for (int i = 0; i < 10; ++i) {
            res[digitOrder[i]] = count[keyOrder[i] - 'a'];
            update(count, dict[digitOrder[i]], res[digitOrder[i]]);
        }
        string str = "";
        for (int i = 0; i < 10; ++i) {
            for (int j = 0; j < res[i]; ++j) {
                str += ('0' + i);
            }
        }
        return str;
    }
};