Add python solutions for problem 90, 92

problems/90.subsets-ii.md

@@ -46,7 +46,7 @@ Output:
 
 ## 代码
 
-* 语言支持：JS，C++
+* 语言支持：JS，C++，Python3
 
 JavaScript Code：
 
@@ -132,6 +132,32 @@ public:
     }
 };
 ```
+Python Code:
+
+```Python
+class Solution:
+    def subsetsWithDup(self, nums: List[int], sorted: bool=False) -> List[List[int]]:
+        """回溯法，通过排序参数避免重复排序"""
+        if not nums:
+            return [[]]
+        elif len(nums) == 1:
+            return [[], nums]
+        else:
+            # 先排序，以便去重
+            # 注意，这道题排序花的时间比较多
+            # 因此，增加一个参数，使后续过程不用重复排序，可以大幅提高时间效率
+            if not sorted:
+                nums.sort()
+            # 回溯法
+            pre_lists = self.subsetsWithDup(nums[:-1], sorted=True)
+            all_lists = [i+[nums[-1]] for i in pre_lists] + pre_lists
+            # 去重
+            result = []
+            for i in all_lists:
+                if i not in result:
+                    result.append(i)
+            return result
+```
 
 ## 相关题目
 

problems/92.reverse-linked-list-ii.md

@@ -49,7 +49,7 @@ Output: 1->4->3->2->5->NULL
 
 ## 代码
 
-语言支持：JS, C++
+语言支持：JS, C++, Python3
 
 JavaScript Code:
 
@@ -194,3 +194,53 @@ public:
     }
 };
 ```
+Python Code:
+```Python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
+        """采用先翻转中间部分，之后与不变的前后部分拼接的思路"""
+        # 处理特殊情况
+        if m == n:
+            return head
+
+        # 例行性的先放一个开始节点，方便操作
+        first = ListNode(0)
+        first.next = head
+
+        # 通过以下两个节点记录拼接点
+        before_m = first  # 原链表m前的部分
+        after_n = None  # 原链表n后的部分
+
+        # 通过以下两个节点记录翻转后的链表
+        between_mn_head = None
+        between_mn_end = None
+
+        index = 0
+        cur_node = first
+        while index < n:
+            index += 1
+            cur_node = cur_node.next
+            if index == m - 1:
+                before_m = cur_node
+            elif index == m:
+                between_mn_end = ListNode(cur_node.val)
+                between_mn_head = between_mn_end
+            elif index > m:
+                temp = between_mn_head
+                between_mn_head = ListNode(cur_node.val)
+                between_mn_head.next = temp
+                if index == n:
+                    after_n = cur_node.next
+
+        # 进行拼接
+        between_mn_end.next = after_n
+        before_m.next = between_mn_head
+
+        return first.next
+```