feat: add solutions to lc problems: No.2094~2096

* No.2094.Finding 3-Digit Even Numbers
* No.2095.Delete the Middle Node of a Linked List
* No.2096.Step-By-Step Directions From a Binary Tree Node to Another

Author: yanglbme

solution/2000-2099/2094.Finding 3-Digit Even Numbers/README.md

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+# [2094. 找出 3 位偶数](https://leetcode-cn.com/problems/finding-3-digit-even-numbers)
+
+[English Version](/solution/2000-2099/2094.Finding%203-Digit%20Even%20Numbers/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个整数数组 <code>digits</code> ，其中每个元素是一个数字（<code>0 - 9</code>）。数组中可能存在重复元素。</p>
+
+<p>你需要找出 <strong>所有</strong> 满足下述条件且 <strong>互不相同</strong> 的整数：</p>
+
+<ul>
+	<li>该整数由 <code>digits</code> 中的三个元素按 <strong>任意</strong> 顺序 <strong>依次连接</strong> 组成。</li>
+	<li>该整数不含 <strong>前导零</strong></li>
+	<li>该整数是一个 <strong>偶数</strong></li>
+</ul>
+
+<p>例如，给定的 <code>digits</code> 是 <code>[1, 2, 3]</code> ，整数 <code>132</code> 和 <code>312</code> 满足上面列出的全部条件。</p>
+
+<p>将找出的所有互不相同的整数按 <strong>递增顺序</strong> 排列，并以数组形式返回<em>。</em></p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>digits = [2,1,3,0]
+<strong>输出：</strong>[102,120,130,132,210,230,302,310,312,320]
+<strong>解释：</strong>
+所有满足题目条件的整数都在输出数组中列出。 
+注意，答案数组中不含有 <strong>奇数</strong> 或带 <strong>前导零</strong> 的整数。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>digits = [2,2,8,8,2]
+<strong>输出：</strong>[222,228,282,288,822,828,882]
+<strong>解释：</strong>
+同样的数字（0 - 9）在构造整数时可以重复多次，重复次数最多与其在 <code>digits</code> 中出现的次数一样。 
+在这个例子中，数字 8 在构造 288、828 和 882 时都重复了两次。 
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>digits = [3,7,5]
+<strong>输出：</strong>[]
+<strong>解释：</strong>
+使用给定的 digits 无法构造偶数。
+</pre>
+
+<p><strong>示例 4：</strong></p>
+
+<pre>
+<strong>输入：</strong>digits = [0,2,0,0]
+<strong>输出：</strong>[200]
+<strong>解释：</strong>
+唯一一个不含 <strong>前导零</strong> 且满足全部条件的整数是 200 。
+</pre>
+
+<p><strong>示例 5：</strong></p>
+
+<pre>
+<strong>输入：</strong>digits = [0,0,0]
+<strong>输出：</strong>[]
+<strong>解释：</strong>
+构造的所有整数都会有 <strong>前导零</strong> 。因此，不存在满足题目条件的整数。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>3 &lt;=&nbsp;digits.length &lt;= 100</code></li>
+	<li><code>0 &lt;= digits[i] &lt;= 9</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
+class Solution:
+    def findEvenNumbers(self, digits: List[int]) -> List[int]:
+        ans = []
+        counter = Counter(digits)
+        for i in range(100, 1000, 2):
+            t = []
+            k = i
+            while k:
+                t.append(k % 10)
+                k //= 10
+            cnt = Counter(t)
+            if all([counter[i] >= cnt[i] for i in range(10)]):
+                ans.append(i)
+        return ans
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public int[] findEvenNumbers(int[] digits) {
+        int[] counter = count(digits);
+        List<Integer> ans = new ArrayList<>();
+        for (int i = 100; i < 1000; i += 2) {
+            int[] t = new int[3];
+            for (int j = 0, k = i; k > 0; ++j) {
+                t[j] = k % 10;
+                k /= 10;
+            }
+            int[] cnt = count(t);
+            if (check(counter, cnt)) {
+                ans.add(i);
+            }
+        }
+        return ans.stream().mapToInt(Integer::valueOf).toArray();
+    }
+
+    private boolean check(int[] cnt1, int[] cnt2) {
+        for (int i = 0; i < 10; ++i) {
+            if (cnt1[i] < cnt2[i]) {
+                return false;
+            }
+        }
+        return true;
+    }
+
+    private int[] count(int[] nums) {
+        int[] counter = new int[10];
+        for (int num : nums) {
+            ++counter[num];
+        }
+        return counter;
+    }
+}
+```
+
+### **TypeScript**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2000-2099/2094.Finding 3-Digit Even Numbers/README_EN.md

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+# [2094. Finding 3-Digit Even Numbers](https://leetcode.com/problems/finding-3-digit-even-numbers)
+
+[中文文档](/solution/2000-2099/2094.Finding%203-Digit%20Even%20Numbers/README.md)
+
+## Description
+
+<p>You are given an integer array <code>digits</code>, where each element is a digit. The array may contain duplicates.</p>
+
+<p>You need to find <strong>all</strong> the <strong>unique</strong> integers that follow the given requirements:</p>
+
+<ul>
+	<li>The integer consists of the <strong>concatenation</strong> of <strong>three</strong> elements from <code>digits</code> in <strong>any</strong> arbitrary order.</li>
+	<li>The integer does not have <strong>leading zeros</strong>.</li>
+	<li>The integer is <strong>even</strong>.</li>
+</ul>
+
+<p>For example, if the given <code>digits</code> were <code>[1, 2, 3]</code>, integers <code>132</code> and <code>312</code> follow the requirements.</p>
+
+<p>Return <em>a <strong>sorted</strong> array of the unique integers.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> digits = [2,1,3,0]
+<strong>Output:</strong> [102,120,130,132,210,230,302,310,312,320]
+<strong>Explanation:</strong> 
+All the possible integers that follow the requirements are in the output array. 
+Notice that there are no <strong>odd</strong> integers or integers with <strong>leading zeros</strong>.</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> digits = [2,2,8,8,2]
+<strong>Output:</strong> [222,228,282,288,822,828,882]
+<strong>Explanation:</strong> 
+The same digit can be used as many times as it appears in <code>digits</code>. 
+In this example, the digit 8 is used twice each time in 288, 828, and 882. 
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> digits = [3,7,5]
+<strong>Output:</strong> []
+<strong>Explanation:</strong> 
+No <strong>even</strong> integers can be formed using the given digits.
+</pre>
+
+<p><strong>Example 4:</strong></p>
+
+<pre>
+<strong>Input:</strong> digits = [0,2,0,0]
+<strong>Output:</strong> [200]
+<strong>Explanation:</strong> 
+The only valid integer that can be formed with three digits and <strong>no leading zeros</strong> is 200.
+</pre>
+
+<p><strong>Example 5:</strong></p>
+
+<pre>
+<strong>Input:</strong> digits = [0,0,0]
+<strong>Output:</strong> []
+<strong>Explanation:</strong> 
+All the integers that can be formed have <strong>leading zeros</strong>. Thus, there are no valid integers.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;=&nbsp;digits.length &lt;= 100</code></li>
+	<li><code>0 &lt;= digits[i] &lt;= 9</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def findEvenNumbers(self, digits: List[int]) -> List[int]:
+        ans = []
+        counter = Counter(digits)
+        for i in range(100, 1000, 2):
+            t = []
+            k = i
+            while k:
+                t.append(k % 10)
+                k //= 10
+            cnt = Counter(t)
+            if all([counter[i] >= cnt[i] for i in range(10)]):
+                ans.append(i)
+        return ans
+```
+
+### **Java**
+
+```java
+class Solution {
+    public int[] findEvenNumbers(int[] digits) {
+        int[] counter = count(digits);
+        List<Integer> ans = new ArrayList<>();
+        for (int i = 100; i < 1000; i += 2) {
+            int[] t = new int[3];
+            for (int j = 0, k = i; k > 0; ++j) {
+                t[j] = k % 10;
+                k /= 10;
+            }
+            int[] cnt = count(t);
+            if (check(counter, cnt)) {
+                ans.add(i);
+            }
+        }
+        return ans.stream().mapToInt(Integer::valueOf).toArray();
+    }
+
+    private boolean check(int[] cnt1, int[] cnt2) {
+        for (int i = 0; i < 10; ++i) {
+            if (cnt1[i] < cnt2[i]) {
+                return false;
+            }
+        }
+        return true;
+    }
+
+    private int[] count(int[] nums) {
+        int[] counter = new int[10];
+        for (int num : nums) {
+            ++counter[num];
+        }
+        return counter;
+    }
+}
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2000-2099/2094.Finding 3-Digit Even Numbers/Solution.java

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+class Solution {
+    public int[] findEvenNumbers(int[] digits) {
+        int[] counter = count(digits);
+        List<Integer> ans = new ArrayList<>();
+        for (int i = 100; i < 1000; i += 2) {
+            int[] t = new int[3];
+            for (int j = 0, k = i; k > 0; ++j) {
+                t[j] = k % 10;
+                k /= 10;
+            }
+            int[] cnt = count(t);
+            if (check(counter, cnt)) {
+                ans.add(i);
+            }
+        }
+        return ans.stream().mapToInt(Integer::valueOf).toArray();
+    }
+
+    private boolean check(int[] cnt1, int[] cnt2) {
+        for (int i = 0; i < 10; ++i) {
+            if (cnt1[i] < cnt2[i]) {
+                return false;
+            }
+        }
+        return true;
+    }
+
+    private int[] count(int[] nums) {
+        int[] counter = new int[10];
+        for (int num : nums) {
+            ++counter[num];
+        }
+        return counter;
+    }
+}

solution/2000-2099/2094.Finding 3-Digit Even Numbers/Solution.py

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+class Solution:
+    def findEvenNumbers(self, digits: List[int]) -> List[int]:
+        ans = []
+        counter = Counter(digits)
+        for i in range(100, 1000, 2):
+            t = []
+            k = i
+            while k:
+                t.append(k % 10)
+                k //= 10
+            cnt = Counter(t)
+            if all([counter[i] >= cnt[i] for i in range(10)]):
+                ans.append(i)
+        return ans