2. 还有一种方法就是在DFS过程中 当有重复元素出现就只对当前这个元素走一起,其他重复元素跳过。参考:http://blog.csdn.net/worldwindjp/article/details/23300545 代码如下: 1publicstaticvoiddfs(int[] S,intstart,intlen, ArrayList<Integer> item,ArrayList<ArrayList<Integer>> res){ 2if(item.size()==l...
Given a set of distinct integers,nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, Ifnums=[1,2,3], a solution is: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ] publicList<List<Integer>> subsets(int[]...
A set of practice note, solution, complexity analysis and test bench to leetcode problem set - leetcode/Subset_II_by_backtracking.drawio at b58bcceb0ea27d0756ad72fb6a64b3b547fae221 · brianchiang-tw/leetcode
2、[Leetcode 90]求含有重复数的子集 Subset IIhttps://www.cnblogs.com/inku/p/9976099.html 3、讲解在这:[Leetcode 216]求给定和的数集合 Combination Sum III 【代码】 classSolution {publicList<List<Integer>> subsets(int[] nums) { List<List<Integer>> ans=newArrayList<>(); List<Integer> tmp=...
mx_idx=dp[mx_idx].second; }returnres; } }; 参考资料: https://discuss.leetcode.com/topic/49580/c-o-n-2-solution-56ms https://discuss.leetcode.com/topic/49456/c-solution-with-explanations/2 LeetCode All in One 题目讲解汇总(持续更新中...)...
4.subset https://leetcode-cn.com/problems/subsets/import java.util.ArrayList; import java.util.List; public class No78_subSets { class Solution { List<List<Integer>> output = new ArrayList(); int n, k; /** * 每一轮,与No46类似。不同之处在于,不要交换,因为交换后的1,2,3和1,3,2...
classSolution{public:vector<int>largestDivisibleSubset(vector<int>& nums){intl = nums.size();if(l==0||l==1)returnnums;std::vector<int>dp(l,0),parent(l,-1);sort(nums.begin(), nums.end());intbegin =0;intmaxdp =0;for(inti = l-1; i >=0; --i) ...
题目地址:https://leetcode.com/problems/largest-divisible-subset/description/ 题目描述: Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: ...
classSolution{public: vector<vector<int>>subsets(vector<int>& nums) {sort(nums.begin(), nums.end());intnum_subset =pow(2, nums.size()); vector<vector<int> >res(num_subset,vector<int>());for(inti =0; i < nums.size(); i++)for(intj =0; j < num_subset; j++)if((j >> ...
后面 5 的部分为什么可以填 true 是因为直接加5就可以使 sum = 5,同时 6 也是 true 是因为 dp[i - 1][6 - 5] == dp[i - 1][1] == true,所以这里可以继承过来。 之后的部分都类推,需要结合代码理解填表的过程。 LeetCode 题目总结