hash表实现单位转换

02 Jul 2024 in Algorithm Pageviews

solve actually problem!

某个公司因为业务扩展，需要到另一个星球做太空项目，由于运转周期与地球不同，不同单位的时长与地球有很大的差异，给定如下星球时长的单位以及换算规则如下：

1 year = 22month, 1month = 31 day, 1day = 24hour, 1hour = 60minute, 1minute = 60second;

根据输入，将其尽可能往较大单位转换；如果转换后为0，则跳过单位不再输出；

用例：

输入： ”489607 minute“

输出：”10month 30 day 7 minute”

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    string UintTransForm(const string &str)
    {
        string number;
        string time;
        stringstream ss(str);
        ss >> number >> time;
        unordered_map<string, int> m_hashMap;
        m_hashMap[time] = atoi(number.c_str());

        vector<string> timeVec = {"second", "minute", "hour", "day", "month", "year"};
        unordeded_map<string, int> standUnit =
            {{"year", 0}, {"month", 22}, {"day", 31}, {"hour", 24}, {"minute", 60}, {"second", 60}};
        unordeded_map<string, int> timeIdx =
            {{"second", 0}, {"minute", 1}, {"hour", 2}, {"day", 3}, {"month", 4}, {"year", 5}};
        
        while (m_hashMap[time] >= standUint[time] && (time != "year")) {
            for (auto i = timeIdx[time]; i < timeVec.size() - 1; i++) {
                if (time == timeVec[i]) {
                    m_hashMap[timeVec[i + 1]] = m_hashMap[time] / standUnit[time];
                    m_hashMap[time] = m_hashMap[time] % standUint[time];
                    time = timeVec[i + 1];
                    break;
                }
            }
        }

        string res;
        for (const auto &[s, num] : m_hashMap) {
            if (num != 0) {
                res += to_string(num) + " " + str + " ";
            }
        }
        res.resize(res.size() - 1);
        return res;
    }
};