📝 docs : add 面试题 01.02. 判定是否互为字符重排.md

Author: GoogTech

Diff for: docs/SUMMARY.md

@@ -138,6 +138,8 @@
     * [1528.重新排列字符串](LeetCode刷题之旅及题目解析/1528.重新排列字符串/1528.重新排列字符串.md)
 
 * [面试题](面试题/README.md)
+    * [面试题01](面试题/面试题01/README.md)
+        * [02.判定是否互为字符重排](面试题/面试题01/02.判定是否互为字符重排/02.判定是否互为字符重排.md)
     * [面试题02](面试题/面试题02/README.md)
         * [02.返回倒数第k个节点](面试题/面试题02/02.返回倒数第k个节点/02.返回倒数第k个节点.md)
         * [03.删除中间节点](面试题/面试题02/03.删除中间节点/03.删除中间节点.md)

Diff for: docs/面试题/面试题01/02.判定是否互为字符重排/02.判定是否互为字符重排.md

@@ -0,0 +1,53 @@
+## 02. 判定是否互为字符重排
+> https://leetcode-cn.com/problems/check-permutation-lcci/
+
+
+### Java
+```java
+/*
+ * @Author: Goog Tech
+ * @Date: 2020-08-29 16:59:28
+ * @LastEditTime: 2020-08-29 16:59:55
+ * @Description: https://leetcode-cn.com/problems/check-permutation-lcci/
+ * @FilePath: \leetcode-googtech\面试题01\#02. 判定是否互为字符重排\Solution.java
+ * @WebSite: https://algorithm.show/
+ */
+
+class Solution {
+    public boolean CheckPermutation(String s1, String s2) {
+        if(s1 == s2) return true;
+        if(s1 == null || s2 == null || s1.length() != s2.length()) return false;
+        char[] c1 = s1.toCharArray();
+        char[] c2 = s2.toCharArray();
+        Arrays.sort(c1);
+        Arrays.sort(c2);
+        for(int i = 0; i < c1.length; i++) {
+            if(c1[i] != c2[i]) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
+```
+
+### Python
+```python
+'''
+Author: Goog Tech
+Date: 2020-08-29 16:59:34
+LastEditTime: 2020-08-29 16:59:48
+Description: https://leetcode-cn.com/problems/check-permutation-lcci/
+FilePath: \leetcode-googtech\面试题01\#02. 判定是否互为字符重排\Solution.py
+WebSite: https://algorithm.show/
+'''
+
+class Solution(object):
+    def CheckPermutation(self, s1, s2):
+        """
+        :type s1: str
+        :type s2: str
+        :rtype: bool
+        """
+        return len(s1) == len(s2) and set(s1) == set(s2)
+```

Diff for: docs/面试题/面试题01/README.md

@@ -0,0 +1,3 @@
+<p style="text-align:center;font-size:25px">
+    📝 面试题01
+</p>