diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md
index 5c635d39..5dc18478 100644
--- a/problems/0349.两个数组的交集.md
+++ b/problems/0349.两个数组的交集.md
@@ -127,7 +127,7 @@ class Solution:
         for num in nums2:
             if num in set1:
                 result_set.add(num) # set1里出现的nums2元素 存放到结果
-        return result_set
+        return list(result_set)
 ```
 
 
diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md
index e3014d00..19f9e128 100644
--- a/problems/1002.查找常用字符.md
+++ b/problems/1002.查找常用字符.md
@@ -169,6 +169,29 @@ class Solution {
     }
 }
 ```
+```python
+class Solution:
+    def commonChars(self, words: List[str]) -> List[str]:
+        if not words: return []
+        result = []
+        hash = [0] * 26 # 用来统计所有字符串里字符出现的最小频率
+        for i, c in enumerate(words[0]):  # 用第一个字符串给hash初始化
+            hash[ord(c) - ord('a')] += 1
+        # 统计除第一个字符串外字符的出现频率
+        for i in range(1, len(words)):
+            hashOtherStr = [0] * 26
+            for j in range(len(words[0])):
+                hashOtherStr[ord(words[i][j]) - ord('a')] += 1
+            # 更新hash，保证hash里统计26个字符在所有字符串里出现的最小次数
+            for k in range(26):
+                hash[k] = min(hash[k], hashOtherStr[k])
+        # 将hash统计的字符次数，转成输出形式
+        for i in range(26):
+            while hash[i] != 0: # 注意这里是while，多个重复的字符
+                result.extend(chr(i + ord('a')))
+                hash[i] -= 1
+        return result
+```
 javaScript
 ```js
 var commonChars = function (words) {
diff --git a/problems/1207.独一无二的出现次数.md b/problems/1207.独一无二的出现次数.md
index c1720430..32aa3c3a 100644
--- a/problems/1207.独一无二的出现次数.md
+++ b/problems/1207.独一无二的出现次数.md
@@ -100,7 +100,21 @@ class Solution {
 ```
 
 Python：
-
+```python
+class Solution:
+    def uniqueOccurrences(self, arr: List[int]) -> bool:
+        count = [0] * 2002
+        for i in range(len(arr)):
+            count[arr[i] + 1000] += 1 # 防止负数作为下标
+        freq = [False] * 1002 # 标记相同频率是否重复出现
+        for i in range(2001):
+            if count[i] > 0:
+                if freq[count[i]] == False:
+                    freq[count[i]] = True
+                else:
+                    return False
+        return True
+```
 Go：
 
 JavaScript：