Given a string date representing a Gregorian calendar date formatted as YYYY-MM-DD, return the day number of the year.
Example 1:
Input: date = "2019-01-09"
Output: 9
Explanation: Given date is the 9th day of the year in 2019.
Example 2:
Input: date = "2019-02-10"
Output: 41
Example 3:
Input: date = "2003-03-01"
Output: 60
Example 4:
Input: date = "2004-03-01"
Output: 61
Constraints:
date.length == 10date[4] == date[7] == '-', and all otherdate[i]'s are digitsdaterepresents a calendar date between Jan 1st, 1900 and Dec 31, 2019.
这道题给了一个 年-月-日 的日期字符串，让我们返回该年已经过了多少天。关于每月各多少天，小学的时候就已经背的滚瓜烂熟了吧，一三五七八十腊，三十一天永不差。唯一有变化的就是二月了，闰年是 29 天，所以这里唯一的难点就是判断闰年了吧。先用个数组列出非闰年各个月的天数，然后分别从给定 date 字符串中提取出年月日，并转为整型数。然后将当前月之前的天数都累加到结果 res 中，接下来判断当前月是否大于2月，没有的话都不用判断闰年了。超过了2月就要判断当前年是否是闰年，判断方法很简单，若能被 400 整除，一定是闰年，或着不能被 100 整除，但能被4整除的也是闰年。是闰年的话就再多加一天，最后再加上当前的天数返回即可，参见代码如下：
class Solution {
public:
int dayOfYear(string date) {
vector<int> monthDays{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int res = 0, year = stoi(date.substr(0, 4)), month = stoi(date.substr(5, 2)), day = stoi(date.substr(8, 2));
for (int i = 0; i < month - 1; ++i) {
res += monthDays[i];
}
if (month > 2 && (year % 400 == 0 || (year % 100 != 0 && year % 4 == 0))) {
++res;
}
return res + day;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1154
参考资料：
https://leetcode.com/problems/day-of-the-year/
https://leetcode.com/problems/day-of-the-year/discuss/355916/C%2B%2B-Number-of-Days-in-a-Month