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                # 0078. 子集 ## 題目地址(78. 子集) <https://leetcode-cn.com/problems/subsets/> ## 題目描述 ``` <pre class="calibre18">``` 給定一組不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: nums = [1,2,3] 輸出: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [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: ![](https://img.kancloud.cn/46/ec/46ec3382e572355e2109c47284e37e4b_1341x1080.jpg) > 每一層灰色的部分,表示當前有哪些節點是可以選擇的, 紅色部分則是選擇路徑。1,2,3,4,5,6 則分別表示我們的 6 個子集。 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++ JavaScript Code: ``` <pre class="calibre18">``` <span class="hljs-title">/* * @lc app=leetcode id=78 lang=javascript * * [78] Subsets * * https://leetcode.com/problems/subsets/description/ * * algorithms * Medium (51.19%) * Total Accepted: 351.6K * Total Submissions: 674.8K * Testcase Example: '[1,2,3]' * * Given a set of distinct integers, nums, return all possible subsets (the * power set). * * Note: The solution set must not contain duplicate subsets. * * Example: * * * Input: nums = [1,2,3] * Output: * [ * ? [3], * [1], * [2], * [1,2,3], * [1,3], * [2,3], * [1,2], * [] * ] * */</span> <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++) { 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> subsets = <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, <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">public</span>: <span class="hljs-params">vector</span><<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>> subsets(<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>& nums) { <span class="hljs-keyword">auto</span> ret = <span class="hljs-params">vector</span><<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>>(); <span class="hljs-keyword">auto</span> tmp = <span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>(); backtrack(ret, tmp, nums, <span class="hljs-params">0</span>); <span class="hljs-keyword">return</span> ret; } <span class="hljs-function"><span class="hljs-keyword">void</span> <span class="hljs-title">backtrack</span><span class="hljs-params">(<span class="hljs-params">vector</span><<span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>>& <span class="hljs-params">list</span>, <span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>& tempList, <span class="hljs-params">vector</span><<span class="hljs-keyword">int</span>>& nums, <span class="hljs-keyword">int</span> start)</span> </span>{ <span class="hljs-params">list</span>.push_back(tempList); <span class="hljs-keyword">for</span> (<span class="hljs-keyword">auto</span> i = start; i < nums.size(); ++i) { tempList.push_back(nums[i]); backtrack(<span class="hljs-params">list</span>, tempList, nums, i + <span class="hljs-params">1</span>); tempList.pop_back(); } } }; ``` ``` ## 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) - [90.subsets-ii](90.subsets-ii.html) - [113.path-sum-ii](113.path-sum-ii.html) - [131.palindrome-partitioning](131.palindrome-partitioning.html)
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