feat: add solutions to lc problems: No.2864~2866 (#1672)

* No.2864.Maximum Odd Binary Number
* No.2865.Beautiful Towers I
* No.2866.Beautiful Towers II

Author: yanglbme

solution/2800-2899/2864.Maximum Odd Binary Number/README.md

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+# [2864. 最大二进制奇数](https://leetcode.cn/problems/maximum-odd-binary-number)
+
+[English Version](/solution/2800-2899/2864.Maximum%20Odd%20Binary%20Number/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个 <strong>二进制</strong> 字符串 <code>s</code> ，其中至少包含一个 <code>'1'</code> 。</p>
+
+<p>你必须按某种方式 <strong>重新排列</strong> 字符串中的位，使得到的二进制数字是可以由该组合生成的 <strong>最大二进制奇数</strong> 。</p>
+
+<p>以字符串形式，表示并返回可以由给定组合生成的最大二进制奇数。</p>
+
+<p><strong>注意 </strong>返回的结果字符串 <strong>可以</strong> 含前导零。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "010"
+<strong>输出：</strong>"001"
+<strong>解释：</strong>因为字符串 s 中仅有一个 '1' ，其必须出现在最后一位上。所以答案是 "001" 。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "0101"
+<strong>输出：</strong>"1001"
+<strong>解释：</strong>其中一个 '1' 必须出现在最后一位上。而由剩下的数字可以生产的最大数字是 "100" 。所以答案是 "1001" 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> 仅由 <code>'0'</code> 和 <code>'1'</code> 组成</li>
+	<li><code>s</code> 中至少包含一个 <code>'1'</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 maximumOddBinaryNumber(self, s: str) -> str:
+        cnt = s.count("1")
+        return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public String maximumOddBinaryNumber(String s) {
+        int cnt = 0;
+        for (char c : s.toCharArray()) {
+            if (c == '1') {
+                ++cnt;
+            }
+        }
+        return "1".repeat(cnt - 1) + "0".repeat(s.length() - cnt) + "1";
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string maximumOddBinaryNumber(string s) {
+        int cnt = count_if(s.begin(), s.end(), [](char c) { return c == '1'; });
+        string ans;
+        for (int i = 1; i < cnt; ++i) {
+            ans.push_back('1');
+        }
+        for (int i = 0; i < s.size() - cnt; ++i) {
+            ans.push_back('0');
+        }
+        ans.push_back('1');
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func maximumOddBinaryNumber(s string) string {
+	cnt := strings.Count(s, "1")
+	return strings.Repeat("1", cnt-1) + strings.Repeat("0", len(s)-cnt) + "1"
+}
+```
+
+### **TypeScript**
+
+```ts
+function maximumOddBinaryNumber(s: string): string {
+    let cnt = 0;
+    for (const c of s) {
+        cnt += c === '1' ? 1 : 0;
+    }
+    return '1'.repeat(cnt - 1) + '0'.repeat(s.length - cnt) + '1';
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2864.Maximum Odd Binary Number/README_EN.md

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+# [2864. Maximum Odd Binary Number](https://leetcode.com/problems/maximum-odd-binary-number)
+
+[中文文档](/solution/2800-2899/2864.Maximum%20Odd%20Binary%20Number/README.md)
+
+## Description
+
+<p>You are given a <strong>binary</strong> string <code>s</code> that contains at least one <code>&#39;1&#39;</code>.</p>
+
+<p>You have to <strong>rearrange</strong> the bits in such a way that the resulting binary number is the <strong>maximum odd binary number</strong> that can be created from this combination.</p>
+
+<p>Return <em>a string representing the maximum odd binary number that can be created from the given combination.</em></p>
+
+<p><strong>Note </strong>that the resulting string <strong>can</strong> have leading zeros.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;010&quot;
+<strong>Output:</strong> &quot;001&quot;
+<strong>Explanation:</strong> Because there is just one &#39;1&#39;, it must be in the last position. So the answer is &quot;001&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;0101&quot;
+<strong>Output:</strong> &quot;1001&quot;
+<strong>Explanation: </strong>One of the &#39;1&#39;s must be in the last position. The maximum number that can be made with the remaining digits is &quot;100&quot;. So the answer is &quot;1001&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> consists only of <code>&#39;0&#39;</code> and <code>&#39;1&#39;</code>.</li>
+	<li><code>s</code> contains at least one <code>&#39;1&#39;</code>.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+class Solution:
+    def maximumOddBinaryNumber(self, s: str) -> str:
+        cnt = s.count("1")
+        return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"
+```
+
+### **Java**
+
+```java
+class Solution {
+    public String maximumOddBinaryNumber(String s) {
+        int cnt = 0;
+        for (char c : s.toCharArray()) {
+            if (c == '1') {
+                ++cnt;
+            }
+        }
+        return "1".repeat(cnt - 1) + "0".repeat(s.length() - cnt) + "1";
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string maximumOddBinaryNumber(string s) {
+        int cnt = count_if(s.begin(), s.end(), [](char c) { return c == '1'; });
+        string ans;
+        for (int i = 1; i < cnt; ++i) {
+            ans.push_back('1');
+        }
+        for (int i = 0; i < s.size() - cnt; ++i) {
+            ans.push_back('0');
+        }
+        ans.push_back('1');
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func maximumOddBinaryNumber(s string) string {
+	cnt := strings.Count(s, "1")
+	return strings.Repeat("1", cnt-1) + strings.Repeat("0", len(s)-cnt) + "1"
+}
+```
+
+### **TypeScript**
+
+```ts
+function maximumOddBinaryNumber(s: string): string {
+    let cnt = 0;
+    for (const c of s) {
+        cnt += c === '1' ? 1 : 0;
+    }
+    return '1'.repeat(cnt - 1) + '0'.repeat(s.length - cnt) + '1';
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2864.Maximum Odd Binary Number/Solution.cpp

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+class Solution {
+public:
+    string maximumOddBinaryNumber(string s) {
+        int cnt = count_if(s.begin(), s.end(), [](char c) { return c == '1'; });
+        string ans;
+        for (int i = 1; i < cnt; ++i) {
+            ans.push_back('1');
+        }
+        for (int i = 0; i < s.size() - cnt; ++i) {
+            ans.push_back('0');
+        }
+        ans.push_back('1');
+        return ans;
+    }
+};

solution/2800-2899/2864.Maximum Odd Binary Number/Solution.go

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+func maximumOddBinaryNumber(s string) string {
+	cnt := strings.Count(s, "1")
+	return strings.Repeat("1", cnt-1) + strings.Repeat("0", len(s)-cnt) + "1"
+}

solution/2800-2899/2864.Maximum Odd Binary Number/Solution.java

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+class Solution {
+    public String maximumOddBinaryNumber(String s) {
+        int cnt = 0;
+        for (char c : s.toCharArray()) {
+            if (c == '1') {
+                ++cnt;
+            }
+        }
+        return "1".repeat(cnt - 1) + "0".repeat(s.length() - cnt) + "1";
+    }
+}

solution/2800-2899/2864.Maximum Odd Binary Number/Solution.py

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+class Solution:
+    def maximumOddBinaryNumber(self, s: str) -> str:
+        cnt = s.count("1")
+        return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"

solution/2800-2899/2864.Maximum Odd Binary Number/Solution.ts

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+function maximumOddBinaryNumber(s: string): string {
+    let cnt = 0;
+    for (const c of s) {
+        cnt += c === '1' ? 1 : 0;
+    }
+    return '1'.repeat(cnt - 1) + '0'.repeat(s.length - cnt) + '1';
+}