计算两个日期之间的天数的最有效方法是什么？基本上我在问我们最喜欢的日期时间库是如何实现的。
当我每 4 年运行 1 次迭代时，我很快实现了一个 ~O(n) 的解决方案。（下面附上代码）
我被要求用计算机类解决问题的介绍来实现这一点，但他们只是每天而不是每 4 年迭代一次。所以我不满足于那个解决方案并想出了下面的一个。但是，是否有更有效的解决方案可用？如果是这样，他们是如何实现的？
#include <iostream>
using namespace std;
#define check_leap(year) ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0)))
#define debug(n) cout << n << endl
int get_days(int month, bool leap){
if (month == 2){
if (leap) return 29;
return 28;
} else if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12){
return 31;
} else {
return 30;
}
}
int days[] = {31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334};
#define days_prior_to_month(n) days[n-2]
int num_days_between(int month1, int day1, int month2, int day2, bool leap){
if (month2 > month1)
return ((days_prior_to_month(month2) - days_prior_to_month(month1+1)) + get_days(month1, leap) - day1 + 1 + day2) + ((leap && month1 <= 2 && 2 <= month2) ? 1 : 0);
else if (month2 == month1)
return day2;
return -1;
}
int main(int argc, char * argv[]){
int year, month, day, year2, month2, day2;
cout << "Year: "; cin >> year;
cout << "Month: "; cin >> month;
cout << "Day: "; cin >> day;
cout << "Year 2: "; cin >> year2;
cout << "Month 2: "; cin >> month2;
cout << "Day 2: "; cin >> day2;
int total = 0;
if (year2 != year){
int leapyears = 0;
total += num_days_between(month, day, 12, 31, check_leap(year));
debug(total);
total += num_days_between(1, 1, month2, day2, check_leap(year2));
debug(total);
int originalyear = year;
year++;
year = year + year % 4;
while (year <= year2-1){
leapyears += check_leap(year) ? 1 : 0;
year += 4;
}
total += leapyears * 366;
debug(total);
total += max(year2 - originalyear - leapyears - 1, 0) * 365;
debug(total);
} else {
total = num_days_between(month, day, month2, day2, check_leap(year));
}
cout << "Total Number of Days In Between: " << total << endl;
system("PAUSE");
return 0;
}