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], [] ]
回溯、位操作
方法一：
位向量法：深搜，时间复杂度O(n^2)，空间复杂度O(n)
设置一个位向量组selected，每个元素只有两种状态：选或者不选
1 void subsets(const vector<int> &S, vector<bool> &selected, int step, vector<vector<int>> &result); 2 vector<vector<int>> subsets(vector<int> &S) 3 { 4 sort(S.begin(),S.end()); 5 6 vector<vector<int>> result; 7 vector<bool> selected(S.size(),false); 8 subsets(S, selected, 0, result); 9 return result; 10 } 11 12 void subsets(const vector<int> &S, vector<bool> &selected, int step, vector<vector<int>> &result) 13 { 14 if(step == S.size()) 15 { 16 vector<int> temp; 17 for(int i=0; i<S.size(); i++) 18 { 19 if(selected[i]==true)//=true与==true 20 temp.push_back(S[i]); 21 } 22 result.push_back(temp); 23 return; 24 } 25 26 //不选S[step] 27 selected[step] = false; 28 subsets(S, selected, step+1, result); 29 //选S[step] 30 selected[step] = true; 31 subsets(S, selected, step+1, result); 32 }
方法二：
二进制法:时间复杂度O(n^2)，空间复杂度O(1)
前提：数组中的元素数不超过int位数。用一个int整数表示位向量，第i位为1；表示选择S[i]为0，表示不选择。
这种方法最巧妙，其实可看做位向量法，只不过更加优化
1 vector<vector<int>> subsets(vector<int> &s) 2 { 3 sort(s.begin(),s.end());//要求有序 4 vector<vector<int>> result; 5 const size_t size = s.size();//数组组成 6 vector<int> temp; 7 8 for(size_t i = 0; i<(1<<size); i++) 9 { 10 for(size_t j = 0; j<size; j++) 11 if(i & (1<<j)) 12 temp.push_back(s[j]); 13 result.push_back(temp); 14 temp.clear(); 15 16 } 17 return result; 18 }
完整代码：
1 #include "stdafx.h" 2 #include <vector> 3 #include <iostream> 4 #include <algorithm> 5 6 using namespace std; 7 /* 8 //二进制法:时间复杂度O(n^2)，空间复杂度O(1) 9 vector<vector<int>> subsets(vector<int> &s) 10 { 11 sort(s.begin(),s.end());//要求有序 12 vector<vector<int>> result; 13 const size_t size = s.size();//数组组成 14 vector<int> temp; 15 16 for(size_t i = 0; i<(1<<size); i++) 17 { 18 for(size_t j = 0; j<size; j++) 19 if(i & (1<<j)) 20 temp.push_back(s[j]); 21 result.push_back(temp); 22 temp.clear(); 23 24 } 25 return result; 26 } 27 */ 28 29 //位向量法：深搜，时间复杂度O(n^2)，空间复杂度O(n) 30 //设置一个位向量组，每个元素只有两种状态：选或者不选 31 void subsets(const vector<int> &S, vector<bool> &selected, int step, vector<vector<int>> &result); 32 vector<vector<int>> subsets(vector<int> &S) 33 { 34 sort(S.begin(),S.end()); 35 36 vector<vector<int>> result; 37 vector<bool> selected(S.size(),false); 38 subsets(S, selected, 0, result); 39 return result; 40 } 41 42 void subsets(const vector<int> &S, vector<bool> &selected, int step, vector<vector<int>> &result) 43 { 44 if(step == S.size()) 45 { 46 vector<int> temp; 47 for(int i=0; i<S.size(); i++) 48 { 49 if(selected[i]==true)//=true与==true 50 temp.push_back(S[i]); 51 } 52 result.push_back(temp); 53 return; 54 } 55 56 //不选S[step] 57 selected[step] = false; 58 subsets(S, selected, step+1, result); 59 //选S[step] 60 selected[step] = true; 61 subsets(S, selected, step+1, result); 62 } 63 64 int main() 65 { 66 const int data[6] = {0}; 67 vector<int> s(data,data+1); 68 vector<vector<int>> result = subsets(s); 69 int count = 1; 70 71 for(auto sub : result) 72 { 73 cout << count++ <<" : "; 74 for(auto ch : sub) 75 cout << ch << " "; 76 cout<<endl; 77 } 78 79 return 0; 80 }