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Commit 9e0a1c1

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feat: add solutions to lc problems: No.2864~2866 (doocs#1672)
* No.2864.Maximum Odd Binary Number * No.2865.Beautiful Towers I * No.2866.Beautiful Towers II
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# [2864. 最大二进制奇数](https://leetcode.cn/problems/maximum-odd-binary-number)
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[English Version](/solution/2800-2899/2864.Maximum%20Odd%20Binary%20Number/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 <strong>二进制</strong> 字符串 <code>s</code> ,其中至少包含一个 <code>'1'</code> 。</p>
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<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>
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<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>
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<p><strong>注意 </strong>返回的结果字符串 <strong>可以</strong> 含前导零。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1:</strong></p>
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<pre>
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<strong>输入:</strong>s = "010"
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<strong>输出:</strong>"001"
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<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" 。
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</pre>
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<p><strong class="example">示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>s = "0101"
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<strong>输出:</strong>"1001"
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<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" 。
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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>1 &lt;= s.length &lt;= 100</code></li>
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<li><code>s</code> 仅由 <code>'0'</code> 和 <code>'1'</code> 组成</li>
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<li><code>s</code> 中至少包含一个 <code>'1'</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 maximumOddBinaryNumber(self, s: str) -> str:
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cnt = s.count("1")
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return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"
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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 String maximumOddBinaryNumber(String s) {
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int cnt = 0;
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for (char c : s.toCharArray()) {
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if (c == '1') {
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++cnt;
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}
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}
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return "1".repeat(cnt - 1) + "0".repeat(s.length() - cnt) + "1";
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string maximumOddBinaryNumber(string s) {
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int cnt = count_if(s.begin(), s.end(), [](char c) { return c == '1'; });
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string ans;
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for (int i = 1; i < cnt; ++i) {
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ans.push_back('1');
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}
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for (int i = 0; i < s.size() - cnt; ++i) {
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ans.push_back('0');
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}
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ans.push_back('1');
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maximumOddBinaryNumber(s string) string {
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cnt := strings.Count(s, "1")
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return strings.Repeat("1", cnt-1) + strings.Repeat("0", len(s)-cnt) + "1"
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}
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```
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### **TypeScript**
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```ts
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function maximumOddBinaryNumber(s: string): string {
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let cnt = 0;
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for (const c of s) {
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cnt += c === '1' ? 1 : 0;
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}
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return '1'.repeat(cnt - 1) + '0'.repeat(s.length - cnt) + '1';
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}
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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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# [2864. Maximum Odd Binary Number](https://leetcode.com/problems/maximum-odd-binary-number)
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[中文文档](/solution/2800-2899/2864.Maximum%20Odd%20Binary%20Number/README.md)
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## Description
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<p>You are given a <strong>binary</strong> string <code>s</code> that contains at least one <code>&#39;1&#39;</code>.</p>
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<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>
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<p>Return <em>a string representing the maximum odd binary number that can be created from the given combination.</em></p>
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<p><strong>Note </strong>that the resulting string <strong>can</strong> have leading zeros.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;010&quot;
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<strong>Output:</strong> &quot;001&quot;
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<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;.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;0101&quot;
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<strong>Output:</strong> &quot;1001&quot;
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<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;.
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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>1 &lt;= s.length &lt;= 100</code></li>
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<li><code>s</code> consists only of <code>&#39;0&#39;</code> and <code>&#39;1&#39;</code>.</li>
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<li><code>s</code> contains at least one <code>&#39;1&#39;</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 maximumOddBinaryNumber(self, s: str) -> str:
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cnt = s.count("1")
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return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"
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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 String maximumOddBinaryNumber(String s) {
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int cnt = 0;
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for (char c : s.toCharArray()) {
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if (c == '1') {
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++cnt;
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}
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}
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return "1".repeat(cnt - 1) + "0".repeat(s.length() - cnt) + "1";
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string maximumOddBinaryNumber(string s) {
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int cnt = count_if(s.begin(), s.end(), [](char c) { return c == '1'; });
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string ans;
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for (int i = 1; i < cnt; ++i) {
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ans.push_back('1');
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}
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for (int i = 0; i < s.size() - cnt; ++i) {
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ans.push_back('0');
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}
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ans.push_back('1');
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maximumOddBinaryNumber(s string) string {
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cnt := strings.Count(s, "1")
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return strings.Repeat("1", cnt-1) + strings.Repeat("0", len(s)-cnt) + "1"
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}
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```
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### **TypeScript**
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```ts
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function maximumOddBinaryNumber(s: string): string {
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let cnt = 0;
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for (const c of s) {
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cnt += c === '1' ? 1 : 0;
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}
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return '1'.repeat(cnt - 1) + '0'.repeat(s.length - cnt) + '1';
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}
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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:
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string maximumOddBinaryNumber(string s) {
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int cnt = count_if(s.begin(), s.end(), [](char c) { return c == '1'; });
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string ans;
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for (int i = 1; i < cnt; ++i) {
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ans.push_back('1');
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}
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for (int i = 0; i < s.size() - cnt; ++i) {
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ans.push_back('0');
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}
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ans.push_back('1');
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return ans;
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}
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};
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func maximumOddBinaryNumber(s string) string {
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cnt := strings.Count(s, "1")
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return strings.Repeat("1", cnt-1) + strings.Repeat("0", len(s)-cnt) + "1"
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}
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class Solution {
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public String maximumOddBinaryNumber(String s) {
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int cnt = 0;
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for (char c : s.toCharArray()) {
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if (c == '1') {
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++cnt;
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}
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}
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return "1".repeat(cnt - 1) + "0".repeat(s.length() - cnt) + "1";
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}
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}
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class Solution:
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def maximumOddBinaryNumber(self, s: str) -> str:
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cnt = s.count("1")
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return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"
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function maximumOddBinaryNumber(s: string): string {
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let cnt = 0;
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for (const c of s) {
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cnt += c === '1' ? 1 : 0;
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}
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return '1'.repeat(cnt - 1) + '0'.repeat(s.length - cnt) + '1';
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}

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