zoukankan      html  css  js  c++  java
  • 【Lintcode】017.Subsets

    题目:

    题解:

    Solution 1 ()

    class Solution {
    public:
        vector<vector<int> > subsets(vector<int> &S) {
            vector<vector<int> > res{{}};
            sort(S.begin(), S.end());
            for (int i = 0; i < S.size(); ++i) {
                int size = res.size();
                for (int j = 0; j < size; ++j) {
                    vector<int> instance = res[j];
                    instance.push_back(S[i]);
                    res.push_back(instance);
                    }
            }
            return res;
        }
    };

    Solution 1.2

    class Solution {
    public:
        vector<vector<int> > subsets(vector<int> &S) {
        vector<vector<int> > res(1, vector<int>());
        sort(S.begin(), S.end());
        
        for (int i = 0; i < S.size(); i++) {
            int n = res.size();
            for (int j = 0; j < n; j++) {
                res.push_back(res[j]);
                res.back().push_back(S[i]);
            }
        }
    
        return res;
        }
    };

    Solution 2 ()

    class Solution {
    public:
        vector<vector<int> > subsets(vector<int> &S) {
            vector<vector<int> > res;
            vector<int> v;
            sort(S.begin(), S.end());
            dfs(res, S, v, 0);
            return res;
        }
        void dfs(vector<vector<int> > &res, vector<int> S, vector<int> &v, int pos) {
            res.push_back(v);
            for (int i = pos; i < S.size(); ++i) {
                v.push_back(S[i]);
                dfs(res, S, v, i + 1);
                v.pop_back();
            }
        }
    };

    Bit Manipulation

    This is the most clever solution that I have seen. The idea is that to give all the possible subsets, we just need to exhaust all the possible combinations of the numbers. And each number has only two possibilities: either in or not in a subset. And this can be represented using a bit.

    There is also another a way to visualize this idea. That is, if we use the above example, 1 appears once in every two consecutive subsets, 2 appears twice in every four consecutive subsets, and 3 appears four times in every eight subsets, shown in the following (initially the 8 subsets are all empty):

    [], [], [], [], [], [], [], []

    [], [1], [], [1], [], [1], [], [1]

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

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

    Solution 3 ()

    class Solution {
    public:
        vector<vector<int>> subsets(vector<int>& S) {
            sort(S.begin(), S.end());
            int num_subset = pow(2, S.size()); 
            vector<vector<int> > res(num_subset, vector<int>());
    
            for (int i = 0; i < S.size(); i++) {
                for (int j = 0; j < num_subset; j++) {
                    if ((j >> i) & 1) {
                        res[j].push_back(S[i]);
                    }
                }
            }
            return res;  
        }
    };
  • 相关阅读:
    状态图(Statechart Diagram)—UML图(五) .
    时序图(Sequence Diagram)—UML图(六)
    传说中的WCF(8):玩转消息协定
    SCSF 系列:利用 Smart Client Software Factory 实现 StopLight (Step By Step)
    ASP.NET MVC 3.0(四): 我要MVC潜规则之配置Routing
    传说中的WCF(14):WCF也可以做聊天程序
    部署图(Deployment Diagram)—UML图(九)
    活动图(Activity Diagram)—UML图(四)
    今天做的机试题Socket聊天程序
    UML的基本结构 .
  • 原文地址:https://www.cnblogs.com/Atanisi/p/6869189.html
Copyright © 2011-2022 走看看