题目描述
指定年份 year
和月份 month
,返回 该月天数 。
示例 1:
输入:year = 1992, month = 7
输出:31
示例 2:
输入:year = 2000, month = 2
输出:29
示例 3:
输入:year = 1900, month = 2
输出:28
提示:
1583 <= year <= 2100
1 <= month <= 12
解法
方法一:判断闰年
我们可以先判断给定的年份是否为闰年,如果年份能被 $4$ 整除但不能被 $100$ 整除,或者能被 $400$ 整除,那么这一年就是闰年。
闰年的二月有 $29$ 天,平年的二月有 $28$ 天。
我们可以用一个数组 $days$ 存储当前年份每个月的天数,其中 $days[0]=0$,$days[i]$ 表示当前年份第 $i$ 个月的天数。那么答案就是 $days[month]$。
时间复杂度 $O(1)$,空间复杂度 $O(1)$。
| class Solution:
def numberOfDays(self, year: int, month: int) -> int:
leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
days = [0, 31, 29 if leap else 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
return days[month]
|
| class Solution {
public int numberOfDays(int year, int month) {
boolean leap = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
int[] days = new int[] {0, 31, leap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
return days[month];
}
}
|
| class Solution {
public:
int numberOfDays(int year, int month) {
bool leap = (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0);
vector<int> days = {0, 31, leap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
return days[month];
}
};
|
| func numberOfDays(year int, month int) int {
leap := (year%4 == 0 && year%100 != 0) || (year%400 == 0)
x := 28
if leap {
x = 29
}
days := []int{0, 31, x, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
return days[month]
}
|
| function numberOfDays(year: number, month: number): number {
const leap = (year % 4 === 0 && year % 100 !== 0) || year % 400 === 0;
const days = [0, 31, leap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
return days[month];
}
|