feat: add solutions to lc problem: No.1476. Subrectangle Queries

Author: yanglbme

solution/1400-1499/1476.Subrectangle Queries/README.md

@@ -93,22 +93,72 @@ subrectangleQueries.getValue(2, 2); // 返回 20
 
 <!-- 这里可写通用的实现逻辑 -->
 
+用历史记录列表保存修改历史。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class SubrectangleQueries:
+
+    def __init__(self, rectangle: List[List[int]]):
+        self.rec = rectangle
+        self.history = []
+
+    def updateSubrectangle(self, row1: int, col1: int, row2: int, col2: int, newValue: int) -> None:
+        self.history.append((row1, col1, row2, col2, newValue))
 
+    def getValue(self, row: int, col: int) -> int:
+        for row1, col1, row2, col2, newValue in self.history[::-1]:
+            if row >= row1 and row <= row2 and col >= col1 and col <= col2:
+                return newValue
+        return self.rec[row][col]
+
+
+# Your SubrectangleQueries object will be instantiated and called as such:
+# obj = SubrectangleQueries(rectangle)
+# obj.updateSubrectangle(row1,col1,row2,col2,newValue)
+# param_2 = obj.getValue(row,col)
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class SubrectangleQueries {
+    private int[][] rec;
+    private List<int[]> history;
+
+    public SubrectangleQueries(int[][] rectangle) {
+        rec = rectangle;
+        history = new ArrayList<>();
+    }
+    
+    public void updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) {
+        history.add(new int[]{row1, col1, row2, col2, newValue});
+    }
+    
+    public int getValue(int row, int col) {
+        for (int i = history.size() - 1; i >= 0; --i) {
+            int[] record = history.get(i);
+            if (row >= record[0] && row <= record[2] && col >= record[1] && col <= record[3]) {
+                return record[4];
+            }
+        }
+        return rec[row][col];
+    }
+}
+
+/**
+ * Your SubrectangleQueries object will be instantiated and called as such:
+ * SubrectangleQueries obj = new SubrectangleQueries(rectangle);
+ * obj.updateSubrectangle(row1,col1,row2,col2,newValue);
+ * int param_2 = obj.getValue(row,col);
+ */
 ```
 
 ### **...**

solution/1400-1499/1476.Subrectangle Queries/README_EN.md

@@ -154,18 +154,68 @@ subrectangleQueries.getValue(2, 2); // return 20
 
 ## Solutions
 
+Use history list to save the updated record.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
+class SubrectangleQueries:
+
+    def __init__(self, rectangle: List[List[int]]):
+        self.rec = rectangle
+        self.history = []
+
+    def updateSubrectangle(self, row1: int, col1: int, row2: int, col2: int, newValue: int) -> None:
+        self.history.append((row1, col1, row2, col2, newValue))
 
+    def getValue(self, row: int, col: int) -> int:
+        for row1, col1, row2, col2, newValue in self.history[::-1]:
+            if row >= row1 and row <= row2 and col >= col1 and col <= col2:
+                return newValue
+        return self.rec[row][col]
+
+
+# Your SubrectangleQueries object will be instantiated and called as such:
+# obj = SubrectangleQueries(rectangle)
+# obj.updateSubrectangle(row1,col1,row2,col2,newValue)
+# param_2 = obj.getValue(row,col)
 ```
 
 ### **Java**
 
 ```java
-
+class SubrectangleQueries {
+    private int[][] rec;
+    private List<int[]> history;
+
+    public SubrectangleQueries(int[][] rectangle) {
+        rec = rectangle;
+        history = new ArrayList<>();
+    }
+    
+    public void updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) {
+        history.add(new int[]{row1, col1, row2, col2, newValue});
+    }
+    
+    public int getValue(int row, int col) {
+        for (int i = history.size() - 1; i >= 0; --i) {
+            int[] record = history.get(i);
+            if (row >= record[0] && row <= record[2] && col >= record[1] && col <= record[3]) {
+                return record[4];
+            }
+        }
+        return rec[row][col];
+    }
+}
+
+/**
+ * Your SubrectangleQueries object will be instantiated and called as such:
+ * SubrectangleQueries obj = new SubrectangleQueries(rectangle);
+ * obj.updateSubrectangle(row1,col1,row2,col2,newValue);
+ * int param_2 = obj.getValue(row,col);
+ */
 ```
 
 ### **...**

solution/1400-1499/1476.Subrectangle Queries/Solution.java

@@ -1,29 +1,24 @@
 class SubrectangleQueries {
-
-    int[][] matrix;
+    private int[][] rec;
+    private List<int[]> history;
 
     public SubrectangleQueries(int[][] rectangle) {
-        matrix = new int[rectangle.length][rectangle[0].length];
-        for (int i = 0; i < rectangle.length; i++) {
-            for (int j = 0; j < rectangle[0].length; j++) {
-                matrix[i][j] = rectangle[i][j];
-            }
-        }
+        rec = rectangle;
+        history = new ArrayList<>();
     }
-
+    
     public void updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) {
-        if (row1 > row2 || col1 > col2) {
-            return;
-        }
-        for (int i = row1; i <= row2; i++) {
-            for (int j = col1; j <= col2; j++) {
-                matrix[i][j] = newValue;
-            }
-        }
+        history.add(new int[]{row1, col1, row2, col2, newValue});
     }
-
+    
     public int getValue(int row, int col) {
-        return matrix[row][col];
+        for (int i = history.size() - 1; i >= 0; --i) {
+            int[] record = history.get(i);
+            if (row >= record[0] && row <= record[2] && col >= record[1] && col <= record[3]) {
+                return record[4];
+            }
+        }
+        return rec[row][col];
     }
 }
 

solution/1400-1499/1476.Subrectangle Queries/Solution.py

@@ -0,0 +1,20 @@
+class SubrectangleQueries:
+
+    def __init__(self, rectangle: List[List[int]]):
+        self.rec = rectangle
+        self.history = []
+
+    def updateSubrectangle(self, row1: int, col1: int, row2: int, col2: int, newValue: int) -> None:
+        self.history.append((row1, col1, row2, col2, newValue))
+
+    def getValue(self, row: int, col: int) -> int:
+        for row1, col1, row2, col2, newValue in self.history[::-1]:
+            if row >= row1 and row <= row2 and col >= col1 and col <= col2:
+                return newValue
+        return self.rec[row][col]
+
+
+# Your SubrectangleQueries object will be instantiated and called as such:
+# obj = SubrectangleQueries(rectangle)
+# obj.updateSubrectangle(row1,col1,row2,col2,newValue)
+# param_2 = obj.getValue(row,col)

solution/1900-1999/1907.Count Salary Categories/README.md

@@ -0,0 +1,78 @@
+# [1907. Count Salary Categories](https://leetcode-cn.com/problems/count-salary-categories)
+
+[English Version](/solution/1900-1999/1907.Count%20Salary%20Categories/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Table: <code>Accounts</code></p>
+
+<pre>
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| account_id  | int  |
+| income      | int  |
++-------------+------+
+account_id is the primary key for this table.
+Each row contains information about the monthly income for one bank account.
+</pre>
+
+<p>&nbsp;</p>
+
+<p>Write an SQL query to report the number of bank accounts of each salary category. The salary categories are:</p>
+
+<ul>
+	<li><code>&quot;Low Salary&quot;</code>: All the salaries <strong>strictly less</strong> than <code>$20000</code>.</li>
+	<li><code>&quot;Average Salary&quot;</code>: All the salaries in the <strong>inclusive</strong> range <code>[$20000, $50000]</code>.</li>
+	<li><code>&quot;High Salary&quot;</code>: All the salaries <strong>strictly greater</strong> than <code>$50000</code>.</li>
+</ul>
+
+<p>The result table <strong>must</strong> contain all three categories. If there are no accounts in a category, then report <code>0</code>. Return the result table in <strong>any order</strong>.</p>
+
+<p>The query result format is in the following example.</p>
+
+<p>&nbsp;</p>
+
+<pre>
+Accounts table:
++------------+--------+
+| account_id | income |
++------------+--------+
+| 3          | 108939 |
+| 2          | 12747  |
+| 8          | 87709  |
+| 6          | 91796  |
++------------+--------+
+
+Result table:
++----------------+----------------+
+| category       | accounts_count |
++----------------+----------------+
+| Low Salary     | 1              |
+| Average Salary | 0              |
+| High Salary    | 3              |
++----------------+----------------+
+
+Low Salary: Account 2.
+Average Salary: No accounts.
+High Salary: Accounts 3, 6, and 8.
+</pre>
+
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/1900-1999/1907.Count Salary Categories/README_EN.md

@@ -0,0 +1,72 @@
+# [1907. Count Salary Categories](https://leetcode.com/problems/count-salary-categories)
+
+[中文文档](/solution/1900-1999/1907.Count%20Salary%20Categories/README.md)
+
+## Description
+
+<p>Table: <code>Accounts</code></p>
+
+<pre>
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| account_id  | int  |
+| income      | int  |
++-------------+------+
+account_id is the primary key for this table.
+Each row contains information about the monthly income for one bank account.
+</pre>
+
+<p>&nbsp;</p>
+
+<p>Write an SQL query to report the number of bank accounts of each salary category. The salary categories are:</p>
+
+<ul>
+	<li><code>&quot;Low Salary&quot;</code>: All the salaries <strong>strictly less</strong> than <code>$20000</code>.</li>
+	<li><code>&quot;Average Salary&quot;</code>: All the salaries in the <strong>inclusive</strong> range <code>[$20000, $50000]</code>.</li>
+	<li><code>&quot;High Salary&quot;</code>: All the salaries <strong>strictly greater</strong> than <code>$50000</code>.</li>
+</ul>
+
+<p>The result table <strong>must</strong> contain all three categories. If there are no accounts in a category, then report <code>0</code>. Return the result table in <strong>any order</strong>.</p>
+
+<p>The query result format is in the following example.</p>
+
+<p>&nbsp;</p>
+
+<pre>
+Accounts table:
++------------+--------+
+| account_id | income |
++------------+--------+
+| 3          | 108939 |
+| 2          | 12747  |
+| 8          | 87709  |
+| 6          | 91796  |
++------------+--------+
+
+Result table:
++----------------+----------------+
+| category       | accounts_count |
++----------------+----------------+
+| Low Salary     | 1              |
+| Average Salary | 0              |
+| High Salary    | 3              |
++----------------+----------------+
+
+Low Salary: Account 2.
+Average Salary: No accounts.
+High Salary: Accounts 3, 6, and 8.
+</pre>
+
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **SQL**
+
+```sql
+
+```
+
+<!-- tabs:end -->