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Commit 84ae9d4

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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
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# [2094. 找出 3 位偶数](https://leetcode-cn.com/problems/finding-3-digit-even-numbers)
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[English Version](/solution/2000-2099/2094.Finding%203-Digit%20Even%20Numbers/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<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>
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<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>
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<ul>
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<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>
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<li>该整数不含 <strong>前导零</strong></li>
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<li>该整数是一个 <strong>偶数</strong></li>
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</ul>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>digits = [2,1,3,0]
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<strong>输出:</strong>[102,120,130,132,210,230,302,310,312,320]
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<strong>解释:</strong>
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所有满足题目条件的整数都在输出数组中列出。
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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>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>digits = [2,2,8,8,2]
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<strong>输出:</strong>[222,228,282,288,822,828,882]
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<strong>解释:</strong>
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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。
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在这个例子中,数字 8 在构造 288、828 和 882 时都重复了两次。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>digits = [3,7,5]
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<strong>输出:</strong>[]
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<strong>解释:</strong>
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使用给定的 digits 无法构造偶数。
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</pre>
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<p><strong>示例 4:</strong></p>
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<pre>
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<strong>输入:</strong>digits = [0,2,0,0]
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<strong>输出:</strong>[200]
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<strong>解释:</strong>
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唯一一个不含 <strong>前导零</strong> 且满足全部条件的整数是 200 。
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</pre>
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<p><strong>示例 5:</strong></p>
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<pre>
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<strong>输入:</strong>digits = [0,0,0]
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<strong>输出:</strong>[]
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<strong>解释:</strong>
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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。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>3 &lt;=&nbsp;digits.length &lt;= 100</code></li>
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<li><code>0 &lt;= digits[i] &lt;= 9</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def findEvenNumbers(self, digits: List[int]) -> List[int]:
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ans = []
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counter = Counter(digits)
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for i in range(100, 1000, 2):
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t = []
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k = i
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while k:
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t.append(k % 10)
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k //= 10
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cnt = Counter(t)
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if all([counter[i] >= cnt[i] for i in range(10)]):
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ans.append(i)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] findEvenNumbers(int[] digits) {
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int[] counter = count(digits);
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List<Integer> ans = new ArrayList<>();
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for (int i = 100; i < 1000; i += 2) {
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int[] t = new int[3];
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for (int j = 0, k = i; k > 0; ++j) {
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t[j] = k % 10;
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k /= 10;
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}
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int[] cnt = count(t);
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if (check(counter, cnt)) {
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ans.add(i);
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}
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}
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return ans.stream().mapToInt(Integer::valueOf).toArray();
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}
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private boolean check(int[] cnt1, int[] cnt2) {
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for (int i = 0; i < 10; ++i) {
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if (cnt1[i] < cnt2[i]) {
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return false;
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}
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}
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return true;
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}
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private int[] count(int[] nums) {
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int[] counter = new int[10];
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for (int num : nums) {
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++counter[num];
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}
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return counter;
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}
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}
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```
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### **TypeScript**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2094. Finding 3-Digit Even Numbers](https://leetcode.com/problems/finding-3-digit-even-numbers)
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[中文文档](/solution/2000-2099/2094.Finding%203-Digit%20Even%20Numbers/README.md)
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## Description
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<p>You are given an integer array <code>digits</code>, where each element is a digit. The array may contain duplicates.</p>
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<p>You need to find <strong>all</strong> the <strong>unique</strong> integers that follow the given requirements:</p>
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<ul>
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<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>
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<li>The integer does not have <strong>leading zeros</strong>.</li>
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<li>The integer is <strong>even</strong>.</li>
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</ul>
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<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>
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<p>Return <em>a <strong>sorted</strong> array of the unique integers.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> digits = [2,1,3,0]
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<strong>Output:</strong> [102,120,130,132,210,230,302,310,312,320]
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<strong>Explanation:</strong>
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All the possible integers that follow the requirements are in the output array.
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Notice that there are no <strong>odd</strong> integers or integers with <strong>leading zeros</strong>.</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> digits = [2,2,8,8,2]
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<strong>Output:</strong> [222,228,282,288,822,828,882]
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<strong>Explanation:</strong>
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The same digit can be used as many times as it appears in <code>digits</code>.
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In this example, the digit 8 is used twice each time in 288, 828, and 882.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> digits = [3,7,5]
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<strong>Output:</strong> []
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<strong>Explanation:</strong>
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No <strong>even</strong> integers can be formed using the given digits.
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</pre>
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<p><strong>Example 4:</strong></p>
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<pre>
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<strong>Input:</strong> digits = [0,2,0,0]
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<strong>Output:</strong> [200]
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<strong>Explanation:</strong>
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The only valid integer that can be formed with three digits and <strong>no leading zeros</strong> is 200.
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</pre>
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<p><strong>Example 5:</strong></p>
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<pre>
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<strong>Input:</strong> digits = [0,0,0]
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<strong>Output:</strong> []
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<strong>Explanation:</strong>
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All the integers that can be formed have <strong>leading zeros</strong>. Thus, there are no valid integers.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>3 &lt;=&nbsp;digits.length &lt;= 100</code></li>
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<li><code>0 &lt;= digits[i] &lt;= 9</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def findEvenNumbers(self, digits: List[int]) -> List[int]:
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ans = []
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counter = Counter(digits)
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for i in range(100, 1000, 2):
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t = []
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k = i
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while k:
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t.append(k % 10)
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k //= 10
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cnt = Counter(t)
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if all([counter[i] >= cnt[i] for i in range(10)]):
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ans.append(i)
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int[] findEvenNumbers(int[] digits) {
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int[] counter = count(digits);
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List<Integer> ans = new ArrayList<>();
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for (int i = 100; i < 1000; i += 2) {
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int[] t = new int[3];
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for (int j = 0, k = i; k > 0; ++j) {
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t[j] = k % 10;
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k /= 10;
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}
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int[] cnt = count(t);
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if (check(counter, cnt)) {
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ans.add(i);
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}
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}
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return ans.stream().mapToInt(Integer::valueOf).toArray();
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}
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private boolean check(int[] cnt1, int[] cnt2) {
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for (int i = 0; i < 10; ++i) {
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if (cnt1[i] < cnt2[i]) {
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return false;
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}
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}
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return true;
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}
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private int[] count(int[] nums) {
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int[] counter = new int[10];
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for (int num : nums) {
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++counter[num];
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}
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return counter;
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}
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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class Solution {
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public int[] findEvenNumbers(int[] digits) {
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int[] counter = count(digits);
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List<Integer> ans = new ArrayList<>();
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for (int i = 100; i < 1000; i += 2) {
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int[] t = new int[3];
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for (int j = 0, k = i; k > 0; ++j) {
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t[j] = k % 10;
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k /= 10;
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}
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int[] cnt = count(t);
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if (check(counter, cnt)) {
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ans.add(i);
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}
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}
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return ans.stream().mapToInt(Integer::valueOf).toArray();
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}
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private boolean check(int[] cnt1, int[] cnt2) {
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for (int i = 0; i < 10; ++i) {
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if (cnt1[i] < cnt2[i]) {
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return false;
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}
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}
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return true;
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}
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private int[] count(int[] nums) {
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int[] counter = new int[10];
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for (int num : nums) {
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++counter[num];
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}
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return counter;
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}
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}
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class Solution:
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def findEvenNumbers(self, digits: List[int]) -> List[int]:
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ans = []
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counter = Counter(digits)
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for i in range(100, 1000, 2):
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t = []
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k = i
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while k:
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t.append(k % 10)
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k //= 10
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cnt = Counter(t)
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if all([counter[i] >= cnt[i] for i in range(10)]):
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ans.append(i)
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return ans

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