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feat: add solutions to lc problem: No.1702
No.1702.Maximum Binary String After Change
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‎solution/1700-1799/1702.Maximum Binary String After Change/README.md

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<ul>
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<li>操作 1 ：如果二进制串包含子字符串 <code>"00"</code> ，你可以用 <code>"10"</code> 将其替换。
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<ul>
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<li>比方说， <code>"<strong>00</strong>010" -> "<strong>10</strong>010"</code></li>
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</ul>
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<li>比方说， <code>"000<strong>10</strong>" -> "000<strong>01</strong>"</code></li>
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</ul>
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</li>
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</ul>
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<p>请你返回执行上述操作任意次以后能得到的 <strong>最大二进制字符串</strong> 。如果二进制字符串 <code>x</code> 对应的十进制数字大于二进制字符串 <code>y</code> 对应的十进制数字，那么我们称二进制字符串<em> </em><code>x</code><em> </em>大于二进制字符串<em> </em><code>y</code><em> </em>。</p>
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：脑筋急转弯**
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我们观察发现，操作 2 可以把所有的 $1$ 都移动到字符串的末尾，而操作 1 可以把所有的 `0000..000` 串变为 `111..110`。
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因此，要想得到最大的二进制串，我们应该把所有不在开头的 $1$ 移动到字符串末尾，使得字符串变为 `111..11...00..11` 的形式，然后借助操作 1 把中间的 `000..00` 变为 `111..10`。这样我们最终可以得到一个最多包含一个 $0$ 的二进制字符串，这个字符串就是我们要求的最大二进制串。
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时间复杂度 $O(n)$，忽略答案的空间消耗，空间复杂度 $O(1)$。其中 $n$ 为字符串 `binary` 的长度。
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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 maximumBinaryString(self, binary: str) -> str:
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k = binary.find('0')
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if k == -1:
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return binary
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k += binary[k + 1:].count('0')
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return '1' * k + '0' + '1' * (len(binary) - k - 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 maximumBinaryString(String binary) {
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int k = binary.indexOf('0');
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if (k == -1) {
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return binary;
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}
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int n = binary.length();
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for (int i = k + 1; i < n; ++i) {
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if (binary.charAt(i) == '0') {
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++k;
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}
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}
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char[] ans = binary.toCharArray();
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Arrays.fill(ans, '1');
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ans[k] = '0';
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return String.valueOf(ans);
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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 maximumBinaryString(string binary) {
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int k = binary.find('0');
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if (k == binary.npos) return binary;
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int n = binary.size();
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for (int i = k + 1; i < n; ++i) {
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if (binary[i] == '0') {
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++k;
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}
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}
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return string(k, '1') + '0' + string(n - k - 1, '1');
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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 maximumBinaryString(binary string) string {
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k := strings.IndexByte(binary, '0')
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if k == -1 {
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return binary
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}
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for _, c := range binary[k+1:] {
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if c == '0' {
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k++
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}
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}
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ans := []byte(binary)
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for i := range ans {
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ans[i] = '1'
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}
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ans[k] = '0'
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return string(ans)
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}
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```
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### **...**

‎solution/1700-1799/1702.Maximum Binary String After Change/README_EN.md

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<ul>
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<li>Operation 1: If the number contains the substring <code>&quot;00&quot;</code>, you can replace it with <code>&quot;10&quot;</code>.
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<ul>
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<li>For example, <code>&quot;<u>00</u>010&quot; -&gt; &quot;<u>10</u>010</code>&quot;</li>
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</ul>
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<li>For example, <code>&quot;000<u>10</u>&quot; -&gt; &quot;000<u>01</u>&quot;</code></li>
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</ul>
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</li>
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</ul>
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<p><em>Return the <strong>maximum binary string</strong> you can obtain after any number of operations. Binary string <code>x</code> is greater than binary string <code>y</code> if <code>x</code>&#39;s decimal representation is greater than <code>y</code>&#39;s decimal representation.</em></p>
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### **Python3**
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```python
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class Solution:
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def maximumBinaryString(self, binary: str) -> str:
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k = binary.find('0')
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if k == -1:
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return binary
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k += binary[k + 1:].count('0')
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return '1' * k + '0' + '1' * (len(binary) - k - 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 maximumBinaryString(String binary) {
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int k = binary.indexOf('0');
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if (k == -1) {
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return binary;
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}
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int n = binary.length();
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for (int i = k + 1; i < n; ++i) {
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if (binary.charAt(i) == '0') {
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++k;
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}
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}
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char[] ans = binary.toCharArray();
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Arrays.fill(ans, '1');
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ans[k] = '0';
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return String.valueOf(ans);
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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 maximumBinaryString(string binary) {
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int k = binary.find('0');
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if (k == binary.npos) return binary;
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int n = binary.size();
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for (int i = k + 1; i < n; ++i) {
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if (binary[i] == '0') {
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++k;
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}
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}
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return string(k, '1') + '0' + string(n - k - 1, '1');
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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 maximumBinaryString(binary string) string {
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k := strings.IndexByte(binary, '0')
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if k == -1 {
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return binary
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}
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for _, c := range binary[k+1:] {
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if c == '0' {
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k++
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}
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}
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ans := []byte(binary)
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for i := range ans {
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ans[i] = '1'
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}
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ans[k] = '0'
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return string(ans)
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}
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```
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### **...**

‎solution/1700-1799/1702.Maximum Binary String After Change/Solution.cpp

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class Solution {
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public:
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string maximumBinaryString(string binary) {
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int k = binary.find('0');
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if (k == binary.npos) return binary;
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int n = binary.size();
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for (int i = k + 1; i < n; ++i) {
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if (binary[i] == '0') {
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++k;
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}
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}
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return string(k, '1') + '0' + string(n - k - 1, '1');
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}
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};

‎solution/1700-1799/1702.Maximum Binary String After Change/Solution.go

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func maximumBinaryString(binary string) string {
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k := strings.IndexByte(binary, '0')
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if k == -1 {
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return binary
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}
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for _, c := range binary[k+1:] {
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if c == '0' {
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k++
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}
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}
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ans := []byte(binary)
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for i := range ans {
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ans[i] = '1'
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}
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ans[k] = '0'
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return string(ans)
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}

‎solution/1700-1799/1702.Maximum Binary String After Change/Solution.java

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class Solution {
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public String maximumBinaryString(String binary) {
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int k = binary.indexOf('0');
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if (k == -1) {
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return binary;
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}
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int n = binary.length();
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for (int i = k + 1; i < n; ++i) {
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if (binary.charAt(i) == '0') {
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++k;
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}
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}
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char[] ans = binary.toCharArray();
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Arrays.fill(ans, '1');
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ans[k] = '0';
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return String.valueOf(ans);
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}
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}

‎solution/1700-1799/1702.Maximum Binary String After Change/Solution.py

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class Solution:
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def maximumBinaryString(self, binary: str) -> str:
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k = binary.find('0')
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if k == -1:
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return binary
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k += binary[k + 1 :].count('0')
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return '1' * k + '0' + '1' * (len(binary) - k - 1)

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