[Leetcode] 423. Reconstruct Original Digits from English 解题报告_leetcode 423. reconstruct original digits from eng-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/78394292

本文介绍了一种从乱序的英文数字描述中提取并按升序排列数字的方法。通过分析0-9英文拼写的特点，利用特定字符的唯一性构建哈希映射，进而推算出每个数字的出现次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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"

思路：

我刚刚开始觉得这道题目可以用BackTracking的方法来解决，后来发现想复杂了：其实根据0-9的英文字母的构成方式，可以直接推导出各个字母的出现次数。例如，g只会出现在eight中，x只会出现在six中等等。。。这样当建立起来字符串中字符和出现次数的哈希映射之后，我们就可以直接推导出各个英文单词的出现个数。

代码：

class Solution {
public:
    string originalDigits(string s) {
        map<char, int> letters;
        vector<int> numbers(10, 0);
        string ret;
        for (char c: s) {
            ++letters[c];
        }
        // we have to follow this order strictly
        numbers[8] = letters['g'];
        numbers[6] = letters['x'];
        numbers[4] = letters['u'];
        numbers[2] = letters['w'];
        numbers[0] = letters['z'];
        numbers[5] = letters['f'] - numbers[4];
        numbers[3] = letters['h'] - numbers[8];
        numbers[7] = letters['s'] - numbers[6];
        numbers[1] = letters['o'] - numbers[0] - numbers[2] - numbers[4];
        numbers[9] = letters['i'] - numbers[5] - numbers[6] - numbers[8];
        // construct the result
        for (int i = 0; i < 10; i++) {
            if (numbers[i] > 0) {
                ret.insert(ret.end(), numbers[i], '0' + i);
            }
        }
        return ret;
    }
};