第一题是输入数组的数值不相同,第二题是输入数组的数值有相同的值,第二题在第一题的基础上需要过滤掉那些相同的数值. level代表的是需要进行选择的数值的位置. 78. Subsets 错误解法: class Solution { public: vector<vector<int>> subsets(vector<int>& nums) { vector<vector<int> > result; if(nums.empty()) ret…

Given a set of distinct integers, S, 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 S = [1,2,3], a solution is: [ [3], [1], [2], [1,2,3], [1…

Given a set of distinct integers, nums, return all possible subsets. Note: 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], [] ] 题目标签:Array 方法 1: 题目给了我们一…

题目链接:https://leetcode.com/problems/subsets/#/description   给出一个数组,数组中的元素各不相同,找到该集合的所有子集(包括空集和本身) 举例说明: int []nums={1,2,3}  返回结果如下: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ] 使用回溯法解决上述问题.   相当于对一棵树的深度遍历操作.    上述的输出结果还是有些规律的,也就是按照子集的元素个数来进行分解,…

原题地址 有两种方法: 1. 对于序列S,其子集可以对应为一个二进制数,每一位对应集合中的某个数字,0代表不选,1代表选,比如S={1,2,3},则子集合就是3bit的所有二进制数. 所以,照着二进制位去构造解空间即可. 2. 也可以用DFS做,对于每个元素,要么选,要么不选. 记得先排序,因为结果集的数字要从小到大出现. 代码(DFS版本): vector<vector<int> > res; void dfs(vector<int> &S, vector&l…

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], [] ] 题意: 给定一个含不同整数的集…

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],   [] ]  …

using prev class Solution { List<List<Integer>> res = new ArrayList<List<Integer>>(); public List<List<Integer>> subsets(int[] nums) { List<Integer> temp = new ArrayList<>(); back(temp,nums, 0); return res;…

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…

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set). Note: The solution set must not contain duplicate subsets. Example: Input: [1,2,2] Output: [ [2], [1], [1,2,2], [2,2], [1,2], [] ] 这个题目思路…