题目：原题链接（简单）
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O(1) | O(1) | 32ms (96.43%) |
Ans 2 (Python) | O(1) | O(1) | 40ms (72.62%) |
Ans 3 (Python) |
LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。
解法一：
def dayOfYear(self, date: str) -> int:
year, month, date = date.split("-")
year, month, date = int(year), int(month), int(date)
leap = year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
if leap:
month_days = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
else:
month_days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
return sum(month_days[:month - 1]) + date解法二（优雅化）：
def dayOfYear(self, date: str) -> int:
year, month, date = map(int, date.split("-"))
if year % 4 == 0 and (year % 100 != 0 or year % 400 == 0):
month_days = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
else:
month_days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
return sum(month_days[:month - 1]) + date
