[CareerCup] 9.4 Subsets 子集合

 

9.4 Write a method to return all subsets of a set.

 

LeetCode上的原题，请参见我之前的博客Subsets 子集合和Subsets II 子集合之二。

 

解法一：

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

 

解法二：

class Solution {
public:vector<vector<int> > getSubsets(vector<int> &S) {vector<vector<int> > res;vector<int> out;getSubsetsDFS(S, 0, out, res);return res;}void getSubsetsDFS(vector<int> &S, int pos, vector<int> &out, vector<vector<int> > &res) {res.push_back(out);for (int i = pos; i < S.size(); ++i) {out.push_back(S[i]);getSubsetsDFS(S, i + 1, out ,res);out.pop_back();}}
};

 

解法三：

class Solution {
public:vector<vector<int> > getSubsets(vector<int> &S) {vector<vector<int> > res;int max = 1 << S.size();for (int i = 0; i < max; ++i) {vector<int> out = convertIntToSet(S, i);res.push_back(out);}return res;}vector<int> convertIntToSet(vector<int> &S, int k) {vector<int> sub;int idx = 0;for (int i = k; i > 0; i >>= 1) {if ((i & 1) == 1) {sub.push_back(S[idx]);}++idx;}return sub;}
};