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Author: NinePiece2

Solutions/0090-subsets-ii/README.md

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+<p>Given an integer array <code>nums</code> that may contain duplicates, return <em>all possible</em> <span data-keyword="subset"><em>subsets</em></span><em> (the power set)</em>.</p>
+
+<p>The solution set <strong>must not</strong> contain duplicate subsets. Return the solution in <strong>any order</strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<pre><strong>Input:</strong> nums = [1,2,2]
+<strong>Output:</strong> [[],[1],[1,2],[1,2,2],[2],[2,2]]
+</pre><p><strong class="example">Example 2:</strong></p>
+<pre><strong>Input:</strong> nums = [0]
+<strong>Output:</strong> [[],[0]]
+</pre>
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10</code></li>
+	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
+</ul>

Solutions/0090-subsets-ii/solution.py

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+class Solution:
+    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
+        def dfs(i: int):
+            if i == len(nums):
+                result.append(temp[:])
+                return 
+            temp.append(nums[i])
+            dfs(i + 1)
+            val = temp.pop()
+            while i + 1 < len(nums) and nums[i + 1] == val:
+                i += 1
+
+            dfs(i + 1)
+
+        nums.sort()
+        result = []
+        temp = []
+        dfs(0)
+
+        return result