From 07dd85094c722fa71fb4460781b10a49ebe11efb Mon Sep 17 00:00:00 2001
From: Asterisk <44215173+casnz1601@users.noreply.github.com>
Date: Mon, 11 Oct 2021 16:10:34 +0800
Subject: [PATCH] =?UTF-8?q?Update=200491.=E9=80=92=E5=A2=9E=E5=AD=90?=
 =?UTF-8?q?=E5=BA=8F=E5=88=97.md?=
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补全python代码补充注释
---
 problems/0491.递增子序列.md | 93 ++++++++++++++++++++++++--------
 1 file changed, 71 insertions(+), 22 deletions(-)

diff --git a/problems/0491.递增子序列.md b/problems/0491.递增子序列.md
index e947b2b9..e8ab8eca 100644
--- a/problems/0491.递增子序列.md
+++ b/problems/0491.递增子序列.md
@@ -229,32 +229,81 @@ class Solution {
     }
 }
 ```
-
-
 Python：
+**回溯**
 ```python3
 class Solution:
-    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
-        res = []
-        path = []
-        def backtrack(nums,startIndex):
-            repeat = []  #这里使用数组来进行去重操作
-            if len(path) >=2:
-                res.append(path[:])  #注意这里不要加return，要取树上的节点
-            for i in range(startIndex,len(nums)):
-                if nums[i] in repeat:
-                    continue
-                if len(path) >= 1:
-                    if nums[i] < path[-1]:
-                        continue
-                repeat.append(nums[i])  #记录这个元素在本层用过了，本层后面不能再用了
-                path.append(nums[i])
-                backtrack(nums,i+1)
-                path.pop()
-        backtrack(nums,0)
-        return res
-```
+    def __init__(self):
+        self.paths = []
+        self.path = []
 
+    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
+        '''
+        本题求自增子序列，所以不能改变原数组顺序
+        '''
+        self.backtracking(nums, 0)
+        return self.paths
+
+    def backtracking(self, nums: List[int], start_index: int):
+        # 收集结果，同78.子集，仍要置于终止条件之前
+        if len(self.path) >= 2:
+            # 本题要求所有的节点
+            self.paths.append(self.path[:])
+        
+        # Base Case（可忽略）
+        if start_index == len(nums):
+            return
+
+        # 单层递归逻辑
+        # 深度遍历中每一层都会有一个全新的usage_list用于记录本层元素是否重复使用
+        usage_list = set()
+        # 同层横向遍历
+        for i in range(start_index, len(nums)):
+            # 若当前元素值小于前一个时（非递增）或者曾用过，跳入下一循环
+            if (self.path and nums[i] < self.path[-1]) or nums[i] in usage_list:
+                continue
+            usage_list.add(nums[i])
+            self.path.append(nums[i])
+            self.backtracking(nums, i+1)
+            self.path.pop() 
+```
+**回溯+哈希表去重**
+```python3
+class Solution:
+    def __init__(self):
+        self.paths = []
+        self.path = []
+
+    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
+        '''
+        本题求自增子序列，所以不能改变原数组顺序
+        '''
+        self.backtracking(nums, 0)
+        return self.paths
+
+    def backtracking(self, nums: List[int], start_index: int):
+        # 收集结果，同78.子集，仍要置于终止条件之前
+        if len(self.path) >= 2:
+            # 本题要求所有的节点
+            self.paths.append(self.path[:])
+        
+        # Base Case（可忽略）
+        if start_index == len(nums):
+            return
+
+        # 单层递归逻辑
+        # 深度遍历中每一层都会有一个全新的usage_list用于记录本层元素是否重复使用
+        usage_list = [False] * 201  # 使用列表去重，题中取值范围[-100, 100]
+        # 同层横向遍历
+        for i in range(start_index, len(nums)):
+            # 若当前元素值小于前一个时（非递增）或者曾用过，跳入下一循环
+            if (self.path and nums[i] < self.path[-1]) or usage_list[nums[i]+100] == True:
+                continue
+            usage_list[nums[i]+100] = True
+            self.path.append(nums[i])
+            self.backtracking(nums, i+1)
+            self.path.pop() 
+```
 Go：
 
 ```golang