class Solution {
public:
    int daysBetweenDates(string date1, string date2) {
        
    }
};

I have this function. Suppose I have parameter

date1 = "2020-01-15", date2 = "2019-12-31"

How can i find them as Title? Thank you so much!

CodePudding user response：

Once you parse the date and appropriate month, day, year information is extracted, you can use the <chrono> header to find the difference in days:

#include <chrono>
#include <iostream>

int main() {
    using namespace std::chrono;
    auto tp1 = sys_days{January / 15 / 2020};
    auto tp2 = sys_days{December / 31 / 2019};

    std::cout << (tp1 - tp2) << '\n'; // 15d
}

https://godbolt.org/z/hfM8YWze5

CodePudding user response：

Alternatively, a pre-C 20 solution could be the following:

#include <iostream>
#include <chrono>

int main()
{
    std::string date1 = "2020-01-15";
    std::tm tm1 = {};
    strptime(date1.c_str(), "%Y-%m-%d", &tm1);
    auto tp1 = std::chrono::system_clock::from_time_t(std::mktime(&tm1));
    
    std::string date2 = "2019-12-31";
    std::tm tm2 = {};
    strptime(date2.c_str(), "%Y-%m-%d", &tm2);
    auto tp2 = std::chrono::system_clock::from_time_t(std::mktime(&tm2));
    
    std::chrono::system_clock::duration d = tp1 - tp2;
    using DayLength = std::chrono::duration<int,std::ratio<60*60*24>>;
    DayLength days = std::chrono::duration_cast<DayLength> (d);
    
    std::cout << days.count() << std::endl;

    return 0;
}

I cannot wrap my head enough on how convoluted the date-extraction operation above is, without using C 20 std::chrono::sys_days.

Basically, it makes the following conversions: string -> C string -> struct tm -> time_point -> duration (in periods) -> duration (in days) -> arithmetic type for printing