[LeetCode]78.Subsets

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);
            }//for
            return 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;
            }//if
            for(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]<<" ";
            }//for
            cout<<endl;
        }//for
        return 0;
    }

运行时间