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) | – | – | 36ms (92.68%) |
Ans 2 (Python) | – | – | 36ms (92.68%) |
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（朴素算法）：
def daysBetweenDates(self, date1: str, date2: str) -> int:
if date1 > date2:
date1, date2 = date2, date1
y1, m1, d1 = date1.split("-")
y1, m1, d1 = int(y1), int(m1), int(d1)
y2, m2, d2 = date2.split("-")
y2, m2, d2 = int(y2), int(m2), int(d2)
months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
months_leap = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
ans = 0
# 统计整年
for y in range(y1 + 1, y2):
if y % 4 == 0 and (y % 100 != 0 or y % 400 == 0):
ans += 366
else:
ans += 365
# 统计整月
months1 = months_leap if y1 % 4 == 0 and (y1 % 100 != 0 or y1 % 400 == 0) else months
months2 = months_leap if y1 % 4 == 0 and (y1 % 100 != 0 or y1 % 400 == 0) else months
if y1 != y2:
ans += sum(months1[m1:])
ans += sum(months2[:m2 - 1])
# 统计剩下的日期
if y1 != y2 or m1 != m2:
ans += months1[m1 - 1] - d1
ans += d2
else:
ans += d2 - d1
return ans解法二（使用datetime）：
def daysBetweenDates(self, date1: str, date2: str) -> int:
if date1 > date2:
date1, date2 = date2, date1
y1, m1, d1 = date1.split("-")
y1, m1, d1 = int(y1), int(m1), int(d1)
y2, m2, d2 = date2.split("-")
y2, m2, d2 = int(y2), int(m2), int(d2)
date1 = datetime.datetime(y1, m1, d1)
date2 = datetime.datetime(y2, m2, d2)
return (date2 - date1).days
