计算两个日期之间的天数差C++/java

1--Java

分析:调用java中Calendar类

int days(Date date1,Date date2){
  Calendar cal = new Calendar.getInstance();
  cal.setTime(date1);
  int time1 = cal.get(Calendar.DAY_OF_YEAR);
  cal.setTime(date2);
  int time2 = cal.get(Calendar.DAY_OF_YEAR);
  //long days = Math.abs(time1-time2)/(24*3600*1000);
  return Math.abs(day1-day2);

}

 

2--java

int days(Date date1,Date date2){
  long time1 = Math.abs(date1.getTime()-date2.getTime());
  return (int)time1/(24*3600*1000);

}

 

分析:首先要知道关于闰年的概念,每四年一润,四百年润,且不能被100整除。因为,闰年二月有29天,全年共计366天。

伪代码:首先将日期字符串的年,月,日分开,类型为yyyy-mm-dd,得到两个日期,year1,month1,day1,year2,month2,day2

    如果年份,月份相同:

       return day1-day2;(大减小)

    如果年份相同,月份不同:

        计算date1,该年过了多少天D1;date2,该年过了多少天D2

      return D1-D2;

    如果,年份不同:

      D1=date1 是该年的第多少天

      D2=date2距离该年该年结束还有多少天

      D3=两个年份之间相差的整年数是多少天

      return D1+D2+D

#include <iostream>

using namespace std;

public bool stringToDate(string date,int &year,int &month,int &day){
    year = atoi(date.substr(0,4).c_str());
    month=atio(date.substr(5,2).c_str());
    day=atio(date.substr(8,2).c_str());
    Day[12]={31,29,31,30,31,30,31,31,30,31,30,31};
    if(isLear(year))
      Day[1]=29;
   return year>0&&month>0&&month<=12&&day<=Day[month-1];
}

public bool isLeap(int year){
  return ((year%4==0)||(year%400==0))&&(year&100!=0);
}

public int daysInYear(int year,int month ,int day){

  int days = day;
  int Day[12]={31,28,31,30,31,30,31,31,30,31,30,31};
  if(isLeap(year))
    Day[1]=29;
  for(int i=0;i<month-1;i++){
    days+=Day[i];
  }
  return days;
}

public between_days(string date1,string date2){
  if(stringToDate(date1,int &year1,int &month1,int &day1)&&stringToDate(string date2,int &year2,int &month2,int &day2)){
    if(year1==year2&&month1==month2){
      return day1>day2?day1-day2:day2-day1;
          }

  else if(year1==year2){
    int d1 = daysInYear(year1,month1,day1);
    int d2 = daysInYear(year2,month2,day2);
    return d1>d2?d1-d2:d2-d1;
    }

  else{
    if(year1<year2){
      swap(year1,year2);
      swap(month1,month2);
      swap(day1,day2);
    }
    int D1=0;
    for(int j=year2+1;j<year1;j++){
      if(isLeap(j))
        D1+=366;
      else
        D1+=365;
     }
    int D2=daysInYear(year1,month1,day1);
    if(isLeap(year2))
      D2=D2+366-daysInYear(year2,month2,day2);
    else
      D2=D2+365-daysInYear(year2,month2,day2);
    return D1+D2;
    }
  }
}

 

      

    

 

      

 

posted @ 2017-08-17 21:54 笨鸟多学 阅读(...) 评论(...) 编辑 收藏