diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md
index ab1ef16a..02713c65 100644
--- a/problems/0541.反转字符串II.md
+++ b/problems/0541.反转字符串II.md
@@ -155,34 +155,27 @@ class Solution {
 
 Python：
 ```python
-
-class Solution(object):
-    def reverseStr(self, s, k):
+class Solution:
+    def reverseStr(self, s: str, k: int) -> str:
         """
-        :type s: str
-        :type k: int
-        :rtype: str
+        1. 使用range(start, end, step)来确定需要调换的初始位置
+        2. 对于字符串s = 'abc'，如果使用s[0:999] ===> 'abc'。字符串末尾如果超过最大长度，则会返回至字符串最后一个值，这个特性可以避免一些边界条件的处理。
+        3. 用切片整体替换，而不是一个个替换.
         """
-        from functools import reduce
-        # turn s into a list 
-        s = list(s)
-        
-        # another way to simply use a[::-1], but i feel this is easier to understand
-        def reverse(s):
-            left, right = 0, len(s) - 1
+        def reverse_substring(text):
+            left, right = 0, len(text) - 1
             while left < right:
-                s[left], s[right] = s[right], s[left]
+                text[left], text[right] = text[right], text[left]
                 left += 1
                 right -= 1
-            return s
+            return text
         
-        # make sure we reverse each 2k elements 
-        for i in range(0, len(s), 2*k):
-            s[i:(i+k)] = reverse(s[i:(i+k)])
-        
-        # combine list into str.
-        return reduce(lambda a, b: a+b, s)
+        res = list(s)
+
+        for cur in range(0, len(s), 2 * k):
+            res[cur: cur + k] = reverse_substring(res[cur: cur + k])
         
+        return ''.join(res)
 ```