本文共 990 字，大约阅读时间需要 3 分钟。
题目：（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) O(1) | O ( 1 ) O(1) O(1) | 32ms (96.43%) |
| Ans 2 (Python) | O ( 1 ) O(1) O(1) | O ( 1 ) 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 转载地址：http://uqgdf.baihongyu.com/