Subsets

最后更新于:2022-04-02 01:12:01

# Subsets - 子集 ### Source - leetcode: [Subsets | LeetCode OJ](https://leetcode.com/problems/subsets/) - lintcode: [(17) Subsets](http://www.lintcode.com/en/problem/subsets/) ### Problem Given a set of distinct integers, *nums*, return all possible subsets. #### Note: - Elements in a subset must be in non-descending order. - The solution set must not contain duplicate subsets. For example, If *nums* = `[1,2,3]`, a solution is: ~~~ [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ] ~~~ ### 题解 子集类问题类似Combination,以输入数组`[1, 2, 3]`分析,根据题意,最终返回结果中子集类的元素应该按照升序排列,故首先需要对原数组进行排序。题目的第二点要求是子集不能重复,至此原题即转化为数学中的组合问题。我们首先尝试使用 [DFS](# "Depth-First Search, 深度优先搜索") 进行求解,大致步骤如下: 1. `[1] -> [1, 2] -> [1, 2, 3]` 1. `[2] -> [2, 3]` 1. `[3]` 将上述过程转化为代码即为对数组遍历,每一轮都保存之前的结果并将其依次加入到最终返回结果中。 ### Python ~~~ class Solution: # @param {integer[]} nums # @return {integer[][]} def subsets(self, nums): if nums is None: return [] result = [] nums.sort() self.dfs(nums, 0, [], result) return result def dfs(self, nums, pos, list_temp, ret): # append new object with [] ret.append([] + list_temp) for i in xrange(pos, len(nums)): list_temp.append(nums[i]) self.dfs(nums, i + 1, list_temp, ret) list_temp.pop() ~~~ ### C++ ~~~ class Solution { public: vector> subsets(vector& nums) { vector > result; if (nums.empty()) return result; sort(nums.begin(), nums.end()); vector list; dfs(nums, 0, list, result); return result; } private: void dfs(vector& nums, int pos, vector &list, vector > &ret) { ret.push_back(list); for (int i = pos; i < nums.size(); ++i) { list.push_back(nums[i]); dfs(nums, i + 1, list, ret); list.pop_back(); } } }; ~~~ ### Java ~~~ public class Solution { public List> subsets(int[] nums) { List> result = new ArrayList>(); List list = new ArrayList(); if (nums == null || nums.length == 0) { return result; } Arrays.sort(nums); dfs(nums, 0, list, result); return result; } private void dfs(int[] nums, int pos, List list, List> ret) { // add temp result first ret.add(new ArrayList(list)); for (int i = pos; i < nums.length; i++) { list.add(nums[i]); dfs(nums, i + 1, list, ret); list.remove(list.size() - 1); } } } ~~~ ### 源码分析 Java 和 Python 的代码中在将临时list 添加到最终结果时新生成了对象,(Python 使用`[] +`), 否则最终返回结果将随着`list` 的变化而变化。 **Notice: backTrack(num, i + 1, list, ret);中的『i + 1』不可误写为『pos + 1』,因为`pos`用于每次大的循环,`i`用于内循环,第一次写subsets的时候在这坑了很久... :(** 回溯法可用图示和函数运行的堆栈图来理解,强烈建议**使用图形和递归的思想**分析,以数组`[1, 2, 3]`进行分析。下图所示为`list`及`result`动态变化的过程,箭头向下表示`list.add`及`result.add`操作,箭头向上表示`list.remove`操作。 ![Subsets运行递归调用图](https://docs.gechiui.com/gc-content/uploads/sites/kancloud/2015-10-24_562b1f5a2e2ba.jpg) ### 复杂度分析 对原有数组排序,时间复杂度近似为 O(nlogn)O(n \log n)O(nlogn). 状态数为所有可能的组合数 O(2n)O(2^n)O(2n), 生成每个状态所需的时间复杂度近似为 O(1)O(1)O(1), 如`[1] -> [1, 2]`, 故总的时间复杂度近似为 O(2n)O(2^n)O(2n). 使用了临时空间`list`保存中间结果,`list` 最大长度为数组长度,故空间复杂度近似为 O(n)O(n)O(n). ### Reference - [[NineChap 1.2] Permutation - Woodstock Blog](http://okckd.github.io/blog/2014/06/12/NineChap-Permutation/) - [九章算法 - subsets模板](http://www.jiuzhang.com/solutions/subsets/) - [LeetCode: Subsets 解题报告 - Yu's Garden - 博客园](http://www.cnblogs.com/yuzhangcmu/p/4211815.html)
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