From ecb99c2a296ba503c4d71e59a256b8de0c77db58 Mon Sep 17 00:00:00 2001
From: Asterisk <44215173+casnz1601@users.noreply.github.com>
Date: Mon, 11 Oct 2021 14:44:00 +0800
Subject: [PATCH] =?UTF-8?q?Update=200090.=E5=AD=90=E9=9B=86II.md?=
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修正python代码补充注释
---
 problems/0090.子集II.md | 36 +++++++++++++++++++++++-------------
 1 file changed, 23 insertions(+), 13 deletions(-)

diff --git a/problems/0090.子集II.md b/problems/0090.子集II.md
index 7ae7c6c2..665137e9 100644
--- a/problems/0090.子集II.md
+++ b/problems/0090.子集II.md
@@ -207,20 +207,30 @@ class Solution {
 Python：
 ```python3
 class Solution:
+    def __init__(self):
+        self.paths = []
+        self.path = []
+
     def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
-        res = []  #存放符合条件结果的集合
-        path = []  #用来存放符合条件结果
-        def backtrack(nums,startIndex):
-            res.append(path[:])
-            for i in range(startIndex,len(nums)):
-                if i > startIndex and nums[i] == nums[i - 1]:  #我们要对同一树层使用过的元素进行跳过
-                    continue
-                path.append(nums[i])
-                backtrack(nums,i+1)  #递归
-                path.pop()  #回溯
-        nums = sorted(nums)  #去重需要排序
-        backtrack(nums,0)
-        return res
+        nums.sort()
+        self.backtracking(nums, 0)
+        return self.paths
+
+    def backtracking(self, nums: List[int], start_index: int) -> None:
+        # ps.空集合仍符合要求
+        self.paths.append(self.path[:])
+        # Base Case
+        if start_index == len(nums):
+            return
+        
+        # 单层递归逻辑
+        for i in range(start_index, len(nums)):
+            if i > start_index and nums[i] == nums[i-1]:
+                # 当前后元素值相同时，跳入下一个循环，去重
+                continue
+            self.path.append(nums[i])
+            self.backtracking(nums, i+1)
+            self.path.pop()
 ```
 
 Go：