feat: add solutions to lc problems

.prettierignore

@@ -25,4 +25,5 @@ node_modules/
 /solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
 /solution/1600-1699/1651.Hopper Company Queries III/Solution.sql
 /solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
+/solution/2100-2199/2118.Build the Equation/Solution.sql
 /solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/Solution.sql

solution/0600-0699/0612.Shortest Distance in a Plane/README.md

@@ -65,7 +65,14 @@ Point2D table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT round(sqrt(pow(p1.x - p2.x, 2) + pow(p1.y - p2.y, 2)), 2) AS shortest
+FROM
+    Point2D AS p1,
+    Point2D AS p2
+WHERE p1.x != p2.x OR p1.y != p2.y
+ORDER BY 1
+LIMIT 1;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0612.Shortest Distance in a Plane/README_EN.md

@@ -54,7 +54,14 @@ Point2D table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT round(sqrt(pow(p1.x - p2.x, 2) + pow(p1.y - p2.y, 2)), 2) AS shortest
+FROM
+    Point2D AS p1,
+    Point2D AS p2
+WHERE p1.x != p2.x OR p1.y != p2.y
+ORDER BY 1
+LIMIT 1;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0612.Shortest Distance in a Plane/Solution.sql

@@ -0,0 +1,8 @@
+# Write your MySQL query statement below
+SELECT round(sqrt(pow(p1.x - p2.x, 2) + pow(p1.y - p2.y, 2)), 2) AS shortest
+FROM
+    Point2D AS p1,
+    Point2D AS p2
+WHERE p1.x != p2.x OR p1.y != p2.y
+ORDER BY 1
+LIMIT 1;

solution/2100-2199/2118.Build the Equation/README.md

@@ -109,7 +109,28 @@ Terms 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            power,
+            CASE power
+                WHEN 0 THEN IF(factor > 0, concat('+', factor), factor)
+                WHEN 1 THEN concat(
+                    IF(factor > 0, concat('+', factor), factor),
+                    'X'
+                )
+                ELSE concat(
+                    IF(factor > 0, concat('+', factor), factor),
+                    'X^',
+                    power
+                )
+            END AS it
+        FROM Terms
+    )
+SELECT
+    concat(group_concat(it ORDER BY power DESC SEPARATOR ""), '=0') AS equation
+FROM T;
 ```
 
 <!-- tabs:end -->

solution/2100-2199/2118.Build the Equation/README_EN.md

@@ -101,7 +101,28 @@ Terms table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            power,
+            CASE power
+                WHEN 0 THEN IF(factor > 0, concat('+', factor), factor)
+                WHEN 1 THEN concat(
+                    IF(factor > 0, concat('+', factor), factor),
+                    'X'
+                )
+                ELSE concat(
+                    IF(factor > 0, concat('+', factor), factor),
+                    'X^',
+                    power
+                )
+            END AS it
+        FROM Terms
+    )
+SELECT
+    concat(group_concat(it ORDER BY power DESC SEPARATOR ""), '=0') AS equation
+FROM T;
 ```
 
 <!-- tabs:end -->

solution/2100-2199/2118.Build the Equation/Solution.sql

@@ -0,0 +1,22 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            power,
+            CASE power
+                WHEN 0 THEN IF(factor > 0, concat('+', factor), factor)
+                WHEN 1 THEN concat(
+                    IF(factor > 0, concat('+', factor), factor),
+                    'X'
+                )
+                ELSE concat(
+                    IF(factor > 0, concat('+', factor), factor),
+                    'X^',
+                    power
+                )
+            END AS it
+        FROM Terms
+    )
+SELECT
+    concat(group_concat(it ORDER BY power DESC SEPARATOR ""), '=0') AS equation
+FROM T;

solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md

@@ -39,6 +39,10 @@
 
 **方法一：数学**
 
+假设三个连续的整数分别为 $x-1$, $x$, $x+1$，则它们的和为 $3x$，因此 $num$ 必须是 $3$ 的倍数。如果 $num$ 不是 $3$ 的倍数，则无法表示成三个连续整数的和，返回空数组。否则，令 $x = \frac{num}{3}$，则 $x-1$, $x$, $x+1$ 就是三个连续整数，它们的和为 $num$。
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -48,8 +52,8 @@
 ```python
 class Solution:
     def sumOfThree(self, num: int) -> List[int]:
-        a, b = divmod(num, 3)
-        return [] if b else [a - 1, a, a + 1]
+        x, mod = divmod(num, 3)
+        return [] if mod else [x - 1, x, x + 1]
 ```
 
 ### **Java**
@@ -74,7 +78,9 @@ class Solution {
 class Solution {
 public:
     vector<long long> sumOfThree(long long num) {
-        if (num % 3) return {};
+        if (num % 3) {
+            return {};
+        }
         long long x = num / 3;
         return {x - 1, x, x + 1};
     }
@@ -96,7 +102,13 @@ func sumOfThree(num int64) []int64 {
 ### **TypeScript**
 
 ```ts
-
+function sumOfThree(num: number): number[] {
+    if (num % 3) {
+        return [];
+    }
+    const x = Math.floor(num / 3);
+    return [x - 1, x, x + 1];
+}
 ```
 
 ### **...**

solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md

@@ -40,8 +40,8 @@
 ```python
 class Solution:
     def sumOfThree(self, num: int) -> List[int]:
-        a, b = divmod(num, 3)
-        return [] if b else [a - 1, a, a + 1]
+        x, mod = divmod(num, 3)
+        return [] if mod else [x - 1, x, x + 1]
 ```
 
 ### **Java**
@@ -64,7 +64,9 @@ class Solution {
 class Solution {
 public:
     vector<long long> sumOfThree(long long num) {
-        if (num % 3) return {};
+        if (num % 3) {
+            return {};
+        }
         long long x = num / 3;
         return {x - 1, x, x + 1};
     }
@@ -86,7 +88,13 @@ func sumOfThree(num int64) []int64 {
 ### **TypeScript**
 
 ```ts
-
+function sumOfThree(num: number): number[] {
+    if (num % 3) {
+        return [];
+    }
+    const x = Math.floor(num / 3);
+    return [x - 1, x, x + 1];
+}
 ```
 
 ### **...**

solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.cpp

@@ -1,8 +1,10 @@
-class Solution {
-public:
-    vector<long long> sumOfThree(long long num) {
-        if (num % 3) return {};
-        long long x = num / 3;
-        return {x - 1, x, x + 1};
-    }
+class Solution {
+public:
+    vector<long long> sumOfThree(long long num) {
+        if (num % 3) {
+            return {};
+        }
+        long long x = num / 3;
+        return {x - 1, x, x + 1};
+    }
 };

solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.py

@@ -1,4 +1,4 @@
-class Solution:
-    def sumOfThree(self, num: int) -> List[int]:
-        a, b = divmod(num, 3)
-        return [] if b else [a - 1, a, a + 1]
+class Solution:
+    def sumOfThree(self, num: int) -> List[int]:
+        x, mod = divmod(num, 3)
+        return [] if mod else [x - 1, x, x + 1]

solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.ts

@@ -0,0 +1,7 @@
+function sumOfThree(num: number): number[] {
+    if (num % 3) {
+        return [];
+    }
+    const x = Math.floor(num / 3);
+    return [x - 1, x, x + 1];
+}

solution/2100-2199/2178.Maximum Split of Positive Even Integers/README.md

@@ -59,6 +59,12 @@
 
 **方法一：贪心**
 
+如果 $finalSum$ 是奇数，那么无法拆分成若干个互不相同的正偶数之和，直接返回空数组。
+
+否则，我们可以贪心地按照 $2, 4, 6, \cdots$ 的顺序拆分 $finalSum$，直到 $finalSum$ 无法再拆分出一个不同的正偶数为止，此时我们将剩余的 $finalSum$ 加到最后一个正偶数上即可。
+
+时间复杂度 $O(\sqrt{finalSum})$，忽略答案数组的空间消耗，空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -122,17 +128,16 @@ public:
 ### **Go**
 
 ```go
-func maximumEvenSplit(finalSum int64) []int64 {
-	ans := []int64{}
+func maximumEvenSplit(finalSum int64) (ans []int64) {
 	if finalSum%2 == 1 {
-		return ans
+		return
 	}
 	for i := int64(2); i <= finalSum; i += 2 {
 		ans = append(ans, i)
 		finalSum -= i
 	}
 	ans[len(ans)-1] += finalSum
-	return ans
+	return
 }
 ```
 
@@ -153,6 +158,25 @@ function maximumEvenSplit(finalSum: number): number[] {
 }
 ```
 
+### **C#**
+
+```cs
+public class Solution {
+    public IList<long> MaximumEvenSplit(long finalSum) {
+        IList<long> ans = new List<long>();
+        if (finalSum % 2 == 1) {
+            return ans;
+        }
+        for (long i = 2; i <= finalSum; i += 2) {
+            ans.Add(i);
+            finalSum -= i;
+        }
+        ans[ans.Count - 1] += finalSum;
+        return ans;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/2100-2199/2178.Maximum Split of Positive Even Integers/README_EN.md

@@ -110,17 +110,16 @@ public:
 ### **Go**
 
 ```go
-func maximumEvenSplit(finalSum int64) []int64 {
-	ans := []int64{}
+func maximumEvenSplit(finalSum int64) (ans []int64) {
 	if finalSum%2 == 1 {
-		return ans
+		return
 	}
 	for i := int64(2); i <= finalSum; i += 2 {
 		ans = append(ans, i)
 		finalSum -= i
 	}
 	ans[len(ans)-1] += finalSum
-	return ans
+	return
 }
 ```
 
@@ -141,6 +140,25 @@ function maximumEvenSplit(finalSum: number): number[] {
 }
 ```
 
+### **C#**
+
+```cs
+public class Solution {
+    public IList<long> MaximumEvenSplit(long finalSum) {
+        IList<long> ans = new List<long>();
+        if (finalSum % 2 == 1) {
+            return ans;
+        }
+        for (long i = 2; i <= finalSum; i += 2) {
+            ans.Add(i);
+            finalSum -= i;
+        }
+        ans[ans.Count - 1] += finalSum;
+        return ans;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/2100-2199/2178.Maximum Split of Positive Even Integers/Solution.cs

@@ -0,0 +1,14 @@
+public class Solution {
+    public IList<long> MaximumEvenSplit(long finalSum) {
+        IList<long> ans = new List<long>();
+        if (finalSum % 2 == 1) {
+            return ans;
+        }
+        for (long i = 2; i <= finalSum; i += 2) {
+            ans.Add(i);
+            finalSum -= i;
+        }
+        ans[ans.Count - 1] += finalSum;
+        return ans;
+    }
+}

solution/2100-2199/2178.Maximum Split of Positive Even Integers/Solution.go

@@ -1,12 +1,11 @@
-func maximumEvenSplit(finalSum int64) []int64 {
-	ans := []int64{}
+func maximumEvenSplit(finalSum int64) (ans []int64) {
 	if finalSum%2 == 1 {
-		return ans
+		return
 	}
 	for i := int64(2); i <= finalSum; i += 2 {
 		ans = append(ans, i)
 		finalSum -= i
 	}
 	ans[len(ans)-1] += finalSum
-	return ans
+	return
 }