feat: Add python code for problem 130, 131 (azl397985856#228)

* Fix solution on problem 88 with js, fixs azl397985856#62

* Fix solution on problem 88, fixs azl397985856#62

* Python solution for problem 90

* Add python solution for problem 92

* Add python solution for problem 94

* Add python code for problem 98

* add python code for problem 102

* add python code for problem 113

* Add python solution for problem 130

* Add python code for problem 131

* Add some comments for python code of problem 131

problems/130.surrounded-regions.md

@@ -54,6 +54,8 @@ Surrounded regions shouldn’t be on the border, which means that any 'O' on the
 
 ## 代码
 
+* 语言支持：JS，Python3
+
 ```js
 
 
@@ -104,6 +106,49 @@ var solve = function(board) {
   return board;
 };
 ```
+Python Code：
+```python
+class Solution:
+    def solve(self, board: List[List[str]]) -> None:
+        """
+        Do not return anything, modify board in-place instead.
+        """
+        # 如果数组长或宽小于等于2，则不需要替换
+        if len(board) <= 2 or len(board[0]) <= 2:
+            return
+
+        row, col = len(board), len(board[0])
+
+        def dfs(i, j):
+            """
+            深度优先算法，如果符合条件，替换为A并进一步测试，否则停止
+            """
+            if i < 0 or j < 0 or i >= row or j >= col or board[i][j] != 'O':
+                return
+            board[i][j] = 'A'
+
+            dfs(i - 1, j)
+            dfs(i + 1, j)
+            dfs(i, j - 1)
+            dfs(i, j + 1)
+
+        # 从外围开始
+        for i in range(row):
+            dfs(i, 0)
+            dfs(i, col-1)
+
+        for j in range(col):
+            dfs(0, j)
+            dfs(row-1, j)
+
+        # 最后完成替换
+        for i in range(row):
+            for j in range(col):
+                if board[i][j] == 'O':
+                    board[i][j] = 'X'
+                elif board[i][j] == 'A':
+                    board[i][j] = 'O'
+```
 
 ## 相关题目
 

problems/131.palindrome-partitioning.md

@@ -33,6 +33,8 @@ Output:
 
 ## 代码
 
+* 语言支持：JS，Python3
+
 ```js
 
 
@@ -83,6 +85,28 @@ var partition = function(s) {
 };
 
 
+```
+```python
+class Solution:
+    def partition(self, s: str) -> List[List[str]]:
+        """回溯法"""
+
+        res = []
+
+        def helper(s, tmp):
+            """
+            如果是空字符串，说明已经处理完毕
+            否则逐个字符往前测试，判断是否是回文
+            如果是，则处理剩余字符串，并将已经得到的列表作为参数
+            """
+            if not s:
+                res.append(tmp)
+            for i in range(1, len(s) + 1):
+                if s[:i] == s[:i][::-1]:
+                    helper(s[i:], tmp + [s[:i]])
+
+        helper(s, [])
+        return res
 ```
 
 ## 相关题目