feat: update solutions to lc problem: No.2409

No.2409.Count Days Spent Together

Author: yanglbme

solution/2400-2499/2409.Count Days Spent Together/README.md

@@ -48,9 +48,9 @@
 
 **方法一：模拟**
 
-将日期转换为天数，然后计算两个人在罗马的天数。
+我们将日期转换为天数，然后计算两个人在罗马的天数。
 
-时间复杂度 $O(1)$。
+时间复杂度 $O(C)$，空间复杂度 $O(C)$。其中 $C$ 为常数。
 
 <!-- tabs:start -->
 
@@ -61,14 +61,12 @@
 ```python
 class Solution:
     def countDaysTogether(self, arriveAlice: str, leaveAlice: str, arriveBob: str, leaveBob: str) -> int:
-        if leaveAlice < arriveBob or leaveBob < arriveAlice:
-            return 0
         a = max(arriveAlice, arriveBob)
         b = min(leaveAlice, leaveBob)
-        days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
+        days = (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
         x = sum(days[:int(a[:2]) - 1]) + int(a[3:])
         y = sum(days[:int(b[:2]) - 1]) + int(b[3:])
-        return y - x + 1
+        return max(y - x + 1, 0)
 ```
 
 ### **Java**

solution/2400-2499/2409.Count Days Spent Together/README_EN.md

@@ -47,14 +47,12 @@
 ```python
 class Solution:
     def countDaysTogether(self, arriveAlice: str, leaveAlice: str, arriveBob: str, leaveBob: str) -> int:
-        if leaveAlice < arriveBob or leaveBob < arriveAlice:
-            return 0
         a = max(arriveAlice, arriveBob)
         b = min(leaveAlice, leaveBob)
-        days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
+        days = (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
         x = sum(days[:int(a[:2]) - 1]) + int(a[3:])
         y = sum(days[:int(b[:2]) - 1]) + int(b[3:])
-        return y - x + 1
+        return max(y - x + 1, 0)
 ```
 
 ### **Java**

solution/2400-2499/2409.Count Days Spent Together/Solution.py

@@ -2,11 +2,9 @@ class Solution:
     def countDaysTogether(
         self, arriveAlice: str, leaveAlice: str, arriveBob: str, leaveBob: str
     ) -> int:
-        if leaveAlice < arriveBob or leaveBob < arriveAlice:
-            return 0
         a = max(arriveAlice, arriveBob)
         b = min(leaveAlice, leaveBob)
-        days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
+        days = (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
         x = sum(days[: int(a[:2]) - 1]) + int(a[3:])
         y = sum(days[: int(b[:2]) - 1]) + int(b[3:])
-        return y - x + 1
+        return max(y - x + 1, 0)

solution/2400-2499/2414.Length of the Longest Alphabetical Continuous Substring/README.md

@@ -46,7 +46,7 @@
 
 **方法一：双指针**
 
-利用双指针 $i$ 和 $j$，分别指向当前连续子字符串的起始位置和结束位置。遍历字符串 $s$，如果当前字符 $s[j]$ 比 $s[j-1]$ 大，则 $j$ 向右移动一位，否则更新 $i$ 为 $j$，并更新最长连续子字符串的长度。
+我们用双指针 $i$ 和 $j$ 分别指向当前连续子字符串的起始位置和结束位置。遍历字符串 $s$，如果当前字符 $s[j]$ 比 $s[j-1]$ 大，则 $j$ 向右移动一位，否则更新 $i$ 为 $j$，并更新最长连续子字符串的长度。
 
 时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。