feat: update solutions to lc problems

* No.0741.Cherry Pickup
* No.0796.Rotate String
* No.1463.Cherry Pickup II

Author: yanglbme

README.md

@@ -36,6 +36,7 @@
 
 -   [在排序数组中查找元素的第一个和最后一个位置](/solution/0000-0099/0034.Find%20First%20and%20Last%20Position%20of%20Element%20in%20Sorted%20Array/README.md) - 二分查找
 -   [准时到达的列车最小时速](/solution/1800-1899/1870.Minimum%20Speed%20to%20Arrive%20on%20Time/README.md) - 二分查找
+-   [找到需要补充粉笔的学生编号](/solution/1800-1899/1894.Find%20the%20Student%20that%20Will%20Replace%20the%20Chalk/README.md) - 二分查找
 -   [可移除字符的最大数目](/solution/1800-1899/1898.Maximum%20Number%20of%20Removable%20Characters/README.md) - 二分查找
 
 ### 2. 搜索
@@ -65,6 +66,8 @@
 -   [最小路径和](/solution/0000-0099/0064.Minimum%20Path%20Sum/README.md) - 线性 DP、数字三角形模型
 -   [摘樱桃](/solution/0700-0799/0741.Cherry%20Pickup/README.md) - 线性 DP、数字三角形模型
 -   [摘樱桃 II](/solution/1400-1499/1463.Cherry%20Pickup%20II/README.md) - 线性 DP、数字三角形模型
+-   [最长递增子序列](/solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README.md) - 线性 DP、最长上升子序列模型
+
 
 ### 4. 高级数据结构
 

README_EN.md

@@ -35,6 +35,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
 
 -   [Find First and Last Position of Element in Sorted Array](/solution/0000-0099/0034.Find%20First%20and%20Last%20Position%20of%20Element%20in%20Sorted%20Array/README_EN.md) - Binary search
 -   [Minimum Speed to Arrive on Time](/solution/1800-1899/1870.Minimum%20Speed%20to%20Arrive%20on%20Time/README_EN.md) - Binary search
+-   [Find the Student that Will Replace the Chalk](/solution/1800-1899/1894.Find%20the%20Student%20that%20Will%20Replace%20the%20Chalk/README_EN.md) - Binary search
 -   [Maximum Number of Removable Characters](/solution/1800-1899/1898.Maximum%20Number%20of%20Removable%20Characters/README_EN.md) - Binary search
 
 #### 2. Search
@@ -64,6 +65,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
 -   [Minimum Path Sum](/solution/0000-0099/0064.Minimum%20Path%20Sum/README_EN.md) - Linear problem
 -   [Cherry Pickup](/solution/0700-0799/0741.Cherry%20Pickup/README_EN.md) - Linear problem
 -   [Cherry Pickup II](/solution/1400-1499/1463.Cherry%20Pickup%20II/README_EN.md) - Linear problem
+-   [Longest Increasing Subsequence](/solution/0300-0399/0300.Longest%20Increasing%20Subsequence/README_EN.md) - Linear problem
 
 ### 4. Advanced Data Structures
 

solution/0700-0799/0741.Cherry Pickup/README.md

@@ -52,7 +52,9 @@
 
 **方法一：动态规划**
 
-线性 DP。
+线性 DP。题目中，玩家从 `(0, 0)` 到 `(N-1, N-1)` 后又重新返回到起始点 `(0, 0)`，我们可以视为玩家两次从 `(0, 0)` 出发到 `(N-1, N-1)`。
+
+定义 `dp[k][i1][i2]` 表示两次路径同时走了 k 步，并且第一次走到 `(i1, k-i1)`，第二次走到 `(i2, k-i2)` 的所有路径中，可获得的樱桃数量的最大值。
 
 类似题型：方格取数、传纸条。
 

solution/0700-0799/0796.Rotate String/README.md

@@ -52,9 +52,7 @@
 ```python
 class Solution:
     def rotateString(self, s: str, goal: str) -> bool:
-        if len(s) != len(goal):
-            return False
-        return goal in s + s
+        return len(s) == len(goal) and goal in s + s
 ```
 
 ### **Java**
@@ -64,10 +62,7 @@ class Solution:
 ```java
 class Solution {
     public boolean rotateString(String s, String goal) {
-        if (s.length() != goal.length()) {
-            return false;
-        }
-        return (s + s).contains(goal);
+        return s.length() == goal.length() && (s + s).contains(goal);
     }
 }
 ```
@@ -78,10 +73,7 @@ class Solution {
 class Solution {
 public:
     bool rotateString(string s, string goal) {
-        if (s.size() != goal.size()) {
-            return false;
-        }
-        return !!strstr((s + s).data(), goal.data());
+        return s.size() == goal.size() && strstr((s + s).data(), goal.data());
     }
 };
 ```
@@ -90,10 +82,7 @@ public:
 
 ```go
 func rotateString(s string, goal string) bool {
-	if len(s) != len(goal) {
-		return false
-	}
-	return strings.Contains(s+s, goal)
+	return len(s) == len(goal) && strings.Contains(s+s, goal)
 }
 ```
 

solution/0700-0799/0796.Rotate String/README_EN.md

@@ -37,20 +37,15 @@
 ```python
 class Solution:
     def rotateString(self, s: str, goal: str) -> bool:
-        if len(s) != len(goal):
-            return False
-        return goal in s + s
+        return len(s) == len(goal) and goal in s + s
 ```
 
 ### **Java**
 
 ```java
 class Solution {
     public boolean rotateString(String s, String goal) {
-        if (s.length() != goal.length()) {
-            return false;
-        }
-        return (s + s).contains(goal);
+        return s.length() == goal.length() && (s + s).contains(goal);
     }
 }
 ```
@@ -61,10 +56,7 @@ class Solution {
 class Solution {
 public:
     bool rotateString(string s, string goal) {
-        if (s.size() != goal.size()) {
-            return false;
-        }
-        return !!strstr((s + s).data(), goal.data());
+        return s.size() == goal.size() && strstr((s + s).data(), goal.data());
     }
 };
 ```
@@ -73,10 +65,7 @@ public:
 
 ```go
 func rotateString(s string, goal string) bool {
-	if len(s) != len(goal) {
-		return false
-	}
-	return strings.Contains(s+s, goal)
+	return len(s) == len(goal) && strings.Contains(s+s, goal)
 }
 ```
 

solution/0700-0799/0796.Rotate String/Solution.cpp

@@ -1,9 +1,6 @@
-class Solution {
-public:
-    bool rotateString(string s, string goal) {
-        if (s.size() != goal.size()) {
-            return false;
-        }
-        return !!strstr((s + s).data(), goal.data());
-    }
+class Solution {
+public:
+    bool rotateString(string s, string goal) {
+        return s.size() == goal.size() && strstr((s + s).data(), goal.data());
+    }
 };

solution/0700-0799/0796.Rotate String/Solution.go

@@ -1,6 +1,3 @@
-func rotateString(s string, goal string) bool {
-	if len(s) != len(goal) {
-		return false
-	}
-	return strings.Contains(s+s, goal)
+func rotateString(s string, goal string) bool {
+	return len(s) == len(goal) && strings.Contains(s+s, goal)
 }

solution/0700-0799/0796.Rotate String/Solution.java

@@ -1,8 +1,5 @@
-class Solution {
-    public boolean rotateString(String s, String goal) {
-        if (s.length() != goal.length()) {
-            return false;
-        }
-        return (s + s).contains(goal);
-    }
+class Solution {
+    public boolean rotateString(String s, String goal) {
+        return s.length() == goal.length() && (s + s).contains(goal);
+    }
 }

solution/0700-0799/0796.Rotate String/Solution.py

@@ -1,5 +1,3 @@
-class Solution:
-    def rotateString(self, s: str, goal: str) -> bool:
-        if len(s) != len(goal):
-            return False
-        return goal in s + s
+class Solution:
+    def rotateString(self, s: str, goal: str) -> bool:
+        return len(s) == len(goal) and goal in s + s

solution/1400-1499/1463.Cherry Pickup II/README.md

@@ -75,7 +75,7 @@
 
 **方法一：动态规划**
 
-线性 DP。
+线性 DP。定义 `dp[i][j1][j2]` 表示两个机器人从起始点分别走到坐标 `(i, j1)`, `(i, j2)` 的所有路线中，可获得的樱桃数量的最大值。
 
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