feat: add solutions to lc problem: No.2028

No.2028.Find Missing Observations

Author: yanglbme

solution/2000-2099/2028.Find Missing Observations/README.md

@@ -71,15 +71,41 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
+        m = len(rolls)
+        s = (n + m) * mean - sum(rolls)
+        if s > n * 6 or s < n:
+            return []
+        ans = [s // n] * n
+        for i in range(s % n):
+            ans[i] += 1
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int[] missingRolls(int[] rolls, int mean, int n) {
+        int m = rolls.length;
+        int s = (n + m) * mean;
+        for (int v : rolls) {
+            s -= v;
+        }
+        if (s > n * 6 || s < n) {
+            return new int[0];
+        }
+        int[] ans = new int[n];
+        Arrays.fill(ans, s / n);
+        for (int i = 0; i < s % n; ++i) {
+            ++ans[i];
+        }
+        return ans;
+    }
+}
 ```
 
 ### **TypeScript**
@@ -155,6 +181,46 @@ impl Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> missingRolls(vector<int>& rolls, int mean, int n) {
+        int m = rolls.size();
+        int s = (n + m) * mean;
+        for (int& v : rolls) s -= v;
+        if (s > n * 6 || s < n) return {};
+        vector<int> ans(n, s / n);
+        for (int i = 0; i < s % n; ++i) ++ans[i];
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func missingRolls(rolls []int, mean int, n int) []int {
+	m := len(rolls)
+	s := (n + m) * mean
+	for _, v := range rolls {
+		s -= v
+	}
+	if s > n*6 || s < n {
+		return []int{}
+	}
+	ans := make([]int, n)
+	for i, j := 0, 0; i < n; i, j = i+1, j+1 {
+		ans[i] = s / n
+		if j < s%n {
+			ans[i]++
+		}
+	}
+	return ans
+}
+```
+
 ### **...**
 
 ```

solution/2000-2099/2028.Find Missing Observations/README_EN.md

@@ -55,13 +55,39 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
+        m = len(rolls)
+        s = (n + m) * mean - sum(rolls)
+        if s > n * 6 or s < n:
+            return []
+        ans = [s // n] * n
+        for i in range(s % n):
+            ans[i] += 1
+        return ans
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int[] missingRolls(int[] rolls, int mean, int n) {
+        int m = rolls.length;
+        int s = (n + m) * mean;
+        for (int v : rolls) {
+            s -= v;
+        }
+        if (s > n * 6 || s < n) {
+            return new int[0];
+        }
+        int[] ans = new int[n];
+        Arrays.fill(ans, s / n);
+        for (int i = 0; i < s % n; ++i) {
+            ++ans[i];
+        }
+        return ans;
+    }
+}
 ```
 
 ### **TypeScript**
@@ -137,6 +163,46 @@ impl Solution {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> missingRolls(vector<int>& rolls, int mean, int n) {
+        int m = rolls.size();
+        int s = (n + m) * mean;
+        for (int& v : rolls) s -= v;
+        if (s > n * 6 || s < n) return {};
+        vector<int> ans(n, s / n);
+        for (int i = 0; i < s % n; ++i) ++ans[i];
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func missingRolls(rolls []int, mean int, n int) []int {
+	m := len(rolls)
+	s := (n + m) * mean
+	for _, v := range rolls {
+		s -= v
+	}
+	if s > n*6 || s < n {
+		return []int{}
+	}
+	ans := make([]int, n)
+	for i, j := 0, 0; i < n; i, j = i+1, j+1 {
+		ans[i] = s / n
+		if j < s%n {
+			ans[i]++
+		}
+	}
+	return ans
+}
+```
+
 ### **...**
 
 ```

solution/2000-2099/2028.Find Missing Observations/Solution.cpp

@@ -0,0 +1,12 @@
+class Solution {
+public:
+    vector<int> missingRolls(vector<int>& rolls, int mean, int n) {
+        int m = rolls.size();
+        int s = (n + m) * mean;
+        for (int& v : rolls) s -= v;
+        if (s > n * 6 || s < n) return {};
+        vector<int> ans(n, s / n);
+        for (int i = 0; i < s % n; ++i) ++ans[i];
+        return ans;
+    }
+};

solution/2000-2099/2028.Find Missing Observations/Solution.go

@@ -0,0 +1,18 @@
+func missingRolls(rolls []int, mean int, n int) []int {
+	m := len(rolls)
+	s := (n + m) * mean
+	for _, v := range rolls {
+		s -= v
+	}
+	if s > n*6 || s < n {
+		return []int{}
+	}
+	ans := make([]int, n)
+	for i, j := 0, 0; i < n; i, j = i+1, j+1 {
+		ans[i] = s / n
+		if j < s%n {
+			ans[i]++
+		}
+	}
+	return ans
+}

solution/2000-2099/2028.Find Missing Observations/Solution.java

@@ -0,0 +1,18 @@
+class Solution {
+    public int[] missingRolls(int[] rolls, int mean, int n) {
+        int m = rolls.length;
+        int s = (n + m) * mean;
+        for (int v : rolls) {
+            s -= v;
+        }
+        if (s > n * 6 || s < n) {
+            return new int[0];
+        }
+        int[] ans = new int[n];
+        Arrays.fill(ans, s / n);
+        for (int i = 0; i < s % n; ++i) {
+            ++ans[i];
+        }
+        return ans;
+    }
+}

solution/2000-2099/2028.Find Missing Observations/Solution.py

@@ -0,0 +1,10 @@
+class Solution:
+    def missingRolls(self, rolls: List[int], mean: int, n: int) -> List[int]:
+        m = len(rolls)
+        s = (n + m) * mean - sum(rolls)
+        if s > n * 6 or s < n:
+            return []
+        ans = [s // n] * n
+        for i in range(s % n):
+            ans[i] += 1
+        return ans

solution/2200-2299/2215.Find the Difference of Two Arrays/README.md

@@ -0,0 +1,82 @@
+# [2215. 找出两数组的不同](https://leetcode-cn.com/problems/find-the-difference-of-two-arrays)
+
+[English Version](/solution/2200-2299/2215.Find%20the%20Difference%20of%20Two%20Arrays/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你两个下标从 <code>0</code> 开始的整数数组 <code>nums1</code> 和 <code>nums2</code> ，请你返回一个长度为 <code>2</code> 的列表 <code>answer</code> ，其中：</p>
+
+<ul>
+	<li><code>answer[0]</code> 是 <code>nums1</code> 中所有<strong> 不 </strong>存在于 <code>nums2</code> 中的 <strong>不同</strong> 整数组成的列表。</li>
+	<li><code>answer[1]</code> 是 <code>nums2</code> 中所有<strong> 不 </strong>存在于 <code>nums1</code> 中的 <strong>不同</strong> 整数组成的列表。</li>
+</ul>
+
+<p><strong>注意：</strong>列表中的整数可以按 <strong>任意</strong> 顺序返回。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums1 = [1,2,3], nums2 = [2,4,6]
+<strong>输出：</strong>[[1,3],[4,6]]
+<strong>解释：
+</strong>对于 nums1 ，nums1[1] = 2 出现在 nums2 中下标 0 处，然而 nums1[0] = 1 和 nums1[2] = 3 没有出现在 nums2 中。因此，answer[0] = [1,3]。
+对于 nums2 ，nums2[0] = 2 出现在 nums1 中下标 1 处，然而 nums2[1] = 4 和 nums2[2] = 6 没有出现在 nums2 中。因此，answer[1] = [4,6]。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums1 = [1,2,3,3], nums2 = [1,1,2,2]
+<strong>输出：</strong>[[3],[]]
+<strong>解释：
+</strong>对于 nums1 ，nums1[2] 和 nums1[3] 没有出现在 nums2 中。由于 nums1[2] == nums1[3] ，二者的值只需要在 answer[0] 中出现一次，故 answer[0] = [3]。
+nums2 中的每个整数都在 nums1 中出现，因此，answer[1] = [] 。 
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 1000</code></li>
+	<li><code>-1000 &lt;= nums1[i], nums2[i] &lt;= 1000</code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->