Every few days between LeetCode 5169. date (leap year judgment)

1. Topic

Please write a program to calculate the number of days interval between two dates.

Dates given a string format YYYY-MM-DD, as shown in the example.

示例 1：
输入：date1 = "2019-06-29", date2 = "2019-06-30"
输出：1

示例 2：
输入：date1 = "2020-01-15", date2 = "2019-12-31"
输出：15 

提示：
给定的日期是 1971 年到 2100 年之间的有效日期。

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/number-of-days-between-two-dates
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

2. Problem Solving

• Leap year judgment condition: divisible by 4, but can not be 100, 400 or be divisible
• (year%4 == 0 && year%100 != 0) || year%400 == 0 It is a leap year

class Solution {
	vector<int> month = {0,31,28,31,30,31,30,31,31,30,31,30,31};
public:
    int daysBetweenDates(string date1, string date2) {
    	if(date1 == date2)
    		return 0;
    	if(date1 > date2)
    		swap(date1, date2);
    	int days = 0, y1, m1, d1, y2, m2, d2, i;
    	y1 = (date1[0]-'0')*1000+(date1[1]-'0')*100+(date1[2]-'0')*10+date1[3]-'0';
    	m1 = (date1[5]-'0')*10+date1[6]-'0';
    	d1 = (date1[8]-'0')*10+date1[9]-'0';
    	y2 = (date2[0]-'0')*1000+(date2[1]-'0')*100+(date2[2]-'0')*10+date2[3]-'0';
    	m2 = (date2[5]-'0')*10+date2[6]-'0';
    	d2 = (date2[8]-'0')*10+date2[9]-'0';
    	for(i = 1; i < m1; i++)
    		days -= month[i];
    	if(isleapyear(y1) && m1>2)
    		days--;//闰年2月29天
    	days -= d1;
    	for(i = y1; i < y2; ++i)
    		days += isleapyear(i) ? 366 : 365;
    	for(i = 1; i < m2; i++)
    		days += month[i];
    	if(isleapyear(y2) && m2>2)
    		days++;
    	days += d2;
    	return days;
    }

    bool isleapyear(int& year)
    {
    	if((year%4 == 0 && year%100 != 0) || year%400 == 0)
    		return true;
    	return false;
    }
};