[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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快兩年沒更新博客了，今天看了一下Leetcode，都500多道題了，就隨便交了一道。

開始考慮到用DFS，可是考慮到有些Digit能夠由一個字母惟一確認，而後推算了一下，發現全部Digit均可以肯定，就直接算答案了。很久沒刷過題了，代碼質量直線降低啊。

 

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;
    }
};