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feat: add docsify tabs
添加tabs切换
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3,062 files changed

+102598
-60105
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index.html

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toggleBtnBg: '#42b983'
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}
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},
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tabs: {
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persist : true, // default
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sync : true, // default
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theme : 'classic', // default
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tabComments: true, // default
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tabHeadings: true // default
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},
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plugins: [
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function (hook, vm) {
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hook.beforeEach(function (html) {
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<script src="//unpkg.com/docsify/lib/plugins/emoji.js"></script>
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<script src="//unpkg.com/docsify/lib/plugins/zoom-image.js"></script>
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<script src="//unpkg.com/docsify-dark-mode@0.6.1/dist/index.js"></script>
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<script src="https://cdn.jsdelivr.net/npm/docsify-tabs@1"></script>
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<script src="//unpkg.com/docsify-pagination/dist/docsify-pagination.min.js"></script>
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</body>
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</html>

lcci/01.01.Is Unique/README.md

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<!-- 这里可写通用的实现逻辑 -->
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### Python3
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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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return len(sets) == len(astr)
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```
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### Java
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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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}
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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 -->

lcci/01.01.Is Unique/README_EN.md

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## Solutions
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### Python3
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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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return len(sets) == len(astr)
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```
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### Java
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### **Java**
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```java
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class Solution {
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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 -->

lcci/01.02.Check Permutation/README.md

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<!-- 这里可写通用的实现逻辑 -->
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### Python3
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<!-- tabs:start -->
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### **Python3**
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```python
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return sorted(s1) == sorted(s2)
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```
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### Java
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### **Java**
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}
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```
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<!-- tabs:end -->

lcci/01.02.Check Permutation/README_EN.md

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## Solutions
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### Python3
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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 CheckPermutation(self, s1: str, s2: str) -> bool:
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return sorted(s1) == sorted(s2)
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```
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### **Java**
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class Solution {
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```
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<!-- tabs:end -->

lcci/01.03.String to URL/README.md

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<!-- tabs:start -->
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```python
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return S.replace(' ', '%20')
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```
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### **Java**
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}
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<!-- tabs:end -->

lcci/01.03.String to URL/README_EN.md

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## Solutions
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class Solution:
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return S.replace(' ', '%20')
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```
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class Solution {
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<!-- tabs:end -->

lcci/01.04.Palindrome Permutation/README.md

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<!-- 这里可写通用的实现逻辑 -->
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用哈希表存储每个字符出现的次数。若次数为奇数的字符超过 1 个,则不是回文排列。
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```python
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return cnt <= 1
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```
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<!-- tabs:end -->

lcci/01.04.Palindrome Permutation/README_EN.md

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## Solutions
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class Solution:
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return cnt <= 1
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```
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class Solution {
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lcci/01.05.One Away/README.md

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遍历两个字符串,逐个字符比较判断。
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return self.oneEditAway(first[1:], second[1:])
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```
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lcci/01.05.One Away/README_EN.md

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## Solutions
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class Solution:
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lcci/01.06.Compress String/README.md

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双指针遍历字符串求解。
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return res if len(res) < len(S) else S
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<!-- tabs:end -->

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