# 0090. 子集 II ## 題目地址(90. 子集 II) <https://leetcode-cn.com/problems/subsets-ii/> ## 題目描述 ``` <pre class="calibre18">``` 給定一個可能包含重復元素的整數數組 nums，返回該數組所有可能的子集（冪集）。 說明：解集不能包含重復的子集。 示例: 輸入: [1,2,2] 輸出: [ [2], [1], [1,2,2], [2,2], [1,2], [] ] ``` ``` ## 前置知識 - 回溯 ## 公司 - 阿里 - 騰訊 - 百度 - 字節 ## 思路 這道題目是求集合，并不是`求極值`，因此動態規劃不是特別切合，因此我們需要考慮別的方法。 這種題目其實有一個通用的解法，就是回溯法。 網上也有大神給出了這種回溯法解題的 [通用寫法](https://leetcode.com/problems/combination-sum/discuss/16502/A-general-approach-to-backtracking-questions-in-Java-(Subsets-Permutations-Combination-Sum-Palindrome-Partitioning)>)，這里的所有的解法使用通用方法解答。 除了這道題目還有很多其他題目可以用這種通用解法，具體的題目見后方相關題目部分。 我們先來看下通用解法的解題思路，我畫了一張圖：  > 圖是 [78.subsets](https://github.com/azl397985856/leetcode/blob/master/problems/78.subsets.md)，都差不多，僅做參考。 通用寫法的具體代碼見下方代碼區。 ## 關鍵點解析 - 回溯法 - backtrack 解題公式 ## 代碼 - 語言支持：JS，C++，Python3 JavaScript Code： ``` <pre class="calibre18">``` <span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">backtrack</span>(<span class="hljs-params">list, tempList, nums, start</span>) </span>{ list.push([...tempList]); <span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = start; i < nums.length; i++) { <span class="hljs-title">// 和78.subsets的區別在于這道題nums可以有重復</span> <span class="hljs-title">// 因此需要過濾這種情況</span> <span class="hljs-keyword">if</span> (i > start && nums[i] === nums[i - <span class="hljs-params">1</span>]) <span class="hljs-keyword">continue</span>; tempList.push(nums[i]); backtrack(list, tempList, nums, i + <span class="hljs-params">1</span>); tempList.pop(); } } <span class="hljs-title">/** * @param {number[]} nums * @return {number[][]} */</span> <span class="hljs-keyword">var</span> subsetsWithDup = <span class="hljs-function"><span class="hljs-keyword">function</span> (<span class="hljs-params">nums</span>) </span>{ <span class="hljs-keyword">const</span> list = []; backtrack( list, [], nums.sort((a, b) => a - b), <span class="hljs-params">0</span>, [] ); <span class="hljs-keyword">return</span> list; }; ``` ``` C++ Code： ``` <pre class="calibre18">``` <span class="hljs-keyword">class</span> Solution { <span class="hljs-keyword">private</span>: <span class="hljs-function"><span class="hljs-keyword">void</span> <span class="hljs-title">subsetsWithDup</span><span class="hljs-params">(<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>& nums, size_t start, <span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>& tmp, <span class="hljs-params">vector</span><<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>>& res)</span> </span>{ res.push_back(tmp); <span class="hljs-keyword">for</span> (<span class="hljs-keyword">auto</span> i = start; i < nums.size(); ++i) { <span class="hljs-keyword">if</span> (i > start && nums[i] == nums[i - <span class="hljs-params">1</span>]) <span class="hljs-keyword">continue</span>; tmp.push_back(nums[i]); subsetsWithDup(nums, i + <span class="hljs-params">1</span>, tmp, res); tmp.pop_back(); } } <span class="hljs-keyword">public</span>: <span class="hljs-params">vector</span><<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>> subsetsWithDup(<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>& nums) { <span class="hljs-keyword">auto</span> tmp = <span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>(); <span class="hljs-keyword">auto</span> res = <span class="hljs-params">vector</span><<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>>(); sort(nums.begin(), nums.end()); subsetsWithDup(nums, <span class="hljs-params">0</span>, tmp, res); <span class="hljs-keyword">return</span> res; } }; ``` ``` Python Code: ``` <pre class="calibre18">``` <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">Solution</span>:</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">subsetsWithDup</span><span class="hljs-params">(self, nums: List[int], sorted: bool=False)</span> -> List[List[int]]:</span> <span class="hljs-string">"""回溯法，通過排序參數避免重復排序"""</span> <span class="hljs-keyword">if</span> <span class="hljs-keyword">not</span> nums: <span class="hljs-keyword">return</span> [[]] <span class="hljs-keyword">elif</span> len(nums) == <span class="hljs-params">1</span>: <span class="hljs-keyword">return</span> [[], nums] <span class="hljs-keyword">else</span>: <span class="hljs-title"># 先排序，以便去重</span> <span class="hljs-title"># 注意，這道題排序花的時間比較多</span> <span class="hljs-title"># 因此，增加一個參數，使后續過程不用重復排序，可以大幅提高時間效率</span> <span class="hljs-keyword">if</span> <span class="hljs-keyword">not</span> sorted: nums.sort() <span class="hljs-title"># 回溯法</span> pre_lists = self.subsetsWithDup(nums[:<span class="hljs-params">-1</span>], sorted=<span class="hljs-keyword">True</span>) all_lists = [i+[nums[<span class="hljs-params">-1</span>]] <span class="hljs-keyword">for</span> i <span class="hljs-keyword">in</span> pre_lists] + pre_lists <span class="hljs-title"># 去重</span> result = [] <span class="hljs-keyword">for</span> i <span class="hljs-keyword">in</span> all_lists: <span class="hljs-keyword">if</span> i <span class="hljs-keyword">not</span> <span class="hljs-keyword">in</span> result: result.append(i) <span class="hljs-keyword">return</span> result ``` ``` ## 相關題目 - [39.combination-sum](39.combination-sum.html) - [40.combination-sum-ii](40.combination-sum-ii.html) - [46.permutations](46.permutations.html) - [47.permutations-ii](47.permutations-ii.html) - [78.subsets](78.subsets.html) - [113.path-sum-ii](113.path-sum-ii.html) - [131.palindrome-partitioning](131.palindrome-partitioning.html)