[LeetCode]78.Subsets

2024-02-19 01:38

文章标签 leetcode 78 subsets

本文主要是介绍[LeetCode]78.Subsets，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目

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],
[1,3],
[2,3],
[1,2],
[]
]

分析

充分利用[LeetCode]77.Combinations
本题其实对上题的拓展，上题是一种长度的组合，本题则是穷尽不同长度的组合。
假设S长度为Len，则令 0 <= K <= Len

代码

    /**------------------------------------*   日期：2015-02-06*   作者：SJF0115*   题目: 78.Subsets*   网址：https://oj.leetcode.com/problems/subsets/*   结果：AC*   来源：LeetCode*   博客：---------------------------------------**/#include <iostream>#include <cstring>#include <vector>#include <algorithm>using namespace std;class Solution {public:vector<vector<int> > subsets(vector<int> &S) {int size = S.size();vector<vector<int> > result;vector<int> path;// 排序sort(S.begin(),S.end());// 空集result.push_back(path);// 其他子集for(int i = 1;i <= size;++i){DFS(S,size,i,0,path,result);}//forreturn result;}private:void DFS(vector<int> &s,int n,int k,int index,vector<int> &path,vector<vector<int> > &result){// 一个子集if(path.size() == k){result.push_back(path);return;}//iffor(int i = index;i < n;++i){path.push_back(s[i]);DFS(s,n,k,i+1,path,result);path.pop_back();}//for}};int main(){Solution s;vector<int> num = {4,3,2};vector<vector<int> > result = s.subsets(num);// 输出for(int i = 0;i < result.size();++i){for(int j = 0;j < result[i].size();++j){cout<<result[i][j]<<" ";}//forcout<<endl;}//forreturn 0;}