【Lintcode】017.Subsets

最新推荐文章于 2025-09-05 14:41:09 发布

转载最新推荐文章于 2025-09-05 14:41:09 发布 · 70 阅读

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原文链接：http://www.cnblogs.com/Atanisi/p/6869189.html

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#开发工具

本文介绍了一种生成集合所有可能子集的算法，并提供了三种不同的实现方法：迭代法、递归法及位操作法。通过这些方法，可以有效地解决组合问题中的子集生成任务。

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;  
    }
};