Update 394._decode_string.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/394._decode_string.md

@@ -33,6 +33,8 @@ s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
 
 就很像一个basic calculator，224题
 
+iterative, 栈
+
 beats 100%
 
 ```python
@@ -59,3 +61,77 @@ class Solution:
                 stack[-1][0] += prev_str * cnt
         return stack[0][0]
 ```
+
+
+
+> 思路 2
+******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
+
+递归
+
+beats 100%
+
+```python
+class Solution:
+    def decodeString(self, s):
+        """
+        :type s: str
+        :rtype: str
+        """
+        if not s or len(s) == 0:
+            return ''       
+                    
+        def helper(s):
+            res = ''
+            open_idx = s.find('[')
+            if open_idx != -1: 
+                open_cnt = 1
+                idx = open_idx + 1
+                while idx < len(s) and open_cnt > 0: # 找到与第一个 '[' 匹配的 ']'
+                    if s[idx] == '[':
+                        open_cnt += 1
+                    elif s[idx] == ']':
+                        open_cnt -= 1
+                    idx += 1
+                close_idx = idx - 1
+                tmp_str = ''
+                cnt, i = 0, 0
+                while i < open_idx and not s[i].isdigit(): # 找到数字前可能有的字符串，见test case2
+                    tmp_str += s[i]
+                    i += 1
+                cnt = int(s[i:open_idx])
+                return tmp_str + cnt * helper(s[open_idx+1:close_idx]) + helper(s[close_idx+1:])
+            else:
+                return s
+            
+        return helper(s)
+```
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