leetcode：暴力枚举法之Subsets

leetcode：暴力枚举法之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],
  [2],
  [2,3],
  [1],
  [1,3],
  [1,2],
  [1,2,3]
]

c++实现：

法一递归

#include <iostream>#include <vector>#include <algorithm>using namespace std;void dfs(const vector<int> &S, vector<int> &path, int step,vector<vector<int> > &result);//void dfs(const vector<int> &S, vector<bool> &selected, int step,vector<vector<int> > &result);//递归；增量构造法vector<vector<int> > subsets(vector<int> &S) {sort(S.begin(), S.end()); // 输出要求有序    vector<vector<int> > result;    vector<int> path;    dfs(S, path, 0, result);    return result;}void dfs(const vector<int> &S, vector<int> &path, int step,vector<vector<int> > &result) {if (step == S.size()){result.push_back(path);        return;    }    // 不选S[step]    dfs(S, path, step + 1, result);    // 选S[step]    path.push_back(S[step]);    dfs(S, path, step + 1, result);    path.pop_back();}//递归；位向量法//vector<vector<int> > subsets(vector<int> &S) //{//sort(S.begin(), S.end()); // 输出要求有序//    vector<vector<int> > result;//    vector<bool> selected(S.size(), false);//    dfs(S, selected, 0, result);//    return result;//}//void dfs(const vector<int> &S, vector<bool> &selected, int step,vector<vector<int> > &result) //{//if (step == S.size()) //{//vector<int> subset;//        for (int i = 0; i < S.size(); i++) //{//if (selected[i]) subset.push_back(S[i]);//}//result.push_back(subset);//return;//}//// 不选S[step]//selected[step] = false;//dfs(S, selected, step + 1, result);//// 选S[step]//selected[step] = true;//dfs(S, selected, step + 1, result);//}int main(){int a[3]={1,2,3};vector<int>vec(a,a+3);vector<vector<int>> out;out= subsets(vec);vector<vector<int>>::iterator pp;vector<int>::iterator it;for(pp=out.begin();pp<out.end();pp++){for (it=(*pp).begin();it<(*pp).end();it++){cout<<*it<<" ";}cout<<endl;}  return 0;}

测试结果：

法二迭代：

本方法的前提是：集合的元素不超过int 位数。用一个int 整数表示位向量，第i 位为1，则表示选择S[i]，为0 则不选择。例如S={A,B,C,D}，则0110=6 表示子集{B,C}。这种方法最巧妙。因为它不仅能生成子集，还能方便的表示集合的并、交、差等集合运算。设两个集合的位向量分别为B1 和B2，则B1jB2;B1&B2;B1B2 分别对应集合的并、交、对称差。

二进制法，也可以看做是位向量法，只不过更加优化。

#include <iostream>#include <vector>#include <algorithm>using namespace std;//迭代；二进制法vector<vector<int> > subsets(vector<int> &S) {sort(S.begin(), S.end()); // 输出要求有序vector<vector<int> > result;const size_t n = S.size();vector<int> v;for (size_t i = 0; i < 1 << n; i++){for (size_t j = 0; j < n; j++){if (i & 1 << j)v.push_back(S[j]);}result.push_back(v);v.clear();}return result;} int main(){int a[3]={1,2,3};vector<int>vec(a,a+3);vector<vector<int>> out;out= subsets(vec);vector<vector<int>>::iterator pp;vector<int>::iterator it;for(pp=out.begin();pp<out.end();pp++){for (it=(*pp).begin();it<(*pp).end();it++){cout<<*it<<" ";}cout<<endl;}  return 0;}

测试结果：