leetcode：暴力枚举法之Subsets II

leetcode：暴力枚举法之Subsets II

题目;

Given a collection of integers that might contain duplicates, 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 = [1,2,2], a solution is:
[

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

这题有重复元素,与Subsets类似（没有重复，相当于每个元素只能选0 或1 次），这里扩充到了每个元素可以选0 到若干次而已。

c++实现；

法一：递归

#include <iostream>#include <vector>#include <algorithm>using namespace std;void dfs(const vector<int> &S, vector<int>::iterator start,vector<int> &path, vector<vector<int> > &result) ;void dfs(const vector<int> &S, vector<int> &count,vector<int> &selected, size_t step, vector<vector<int> > &result);//递归；增量构造法//vector<vector<int> > subsetsWithDup(vector<int> &S) //{//sort(S.begin(), S.end()); // 必须排序//    vector<vector<int> > result;//vector<int> path;//dfs(S, S.begin(), path, result);//return result;//}//void dfs(const vector<int> &S, vector<int>::iterator start,vector<int> &path, vector<vector<int> > &result) //{//result.push_back(path);//for (auto i = start; i < S.end(); i++) //{//if (i != start && *i == *(i-1)) //continue;//path.push_back(*i);//dfs(S, i + 1, path, result);//path.pop_back();//}//}//递归；位向量法vector<vector<int> > subsetsWithDup(vector<int> &S) {vector<vector<int> > result; // 必须排序sort(S.begin(), S.end());vector<int> count(S.back() - S.front() + 1, 0);// 计算所有元素的个数for (auto i : S) {count[i - S[0]]++;}// 每个元素选择了多少个vector<int> selected(S.back() - S.front() + 1, -1);dfs(S, count, selected, 0, result);return result;}void dfs(const vector<int> &S, vector<int> &count,vector<int> &selected, size_t step, vector<vector<int> > &result) {if (step == count.size()) {vector<int> subset;for(size_t i = 0; i < selected.size(); i++) {for (int j = 0; j < selected[i]; j++) {subset.push_back(i+S[0]);}}result.push_back(subset);return;}for (int i = 0; i <= count[step]; i++){selected[step] = i;dfs(S, count, selected, step + 1, result);}}int main(){int a[3]={1,2,2};vector<int>vec(a,a+3);vector<vector<int>> out;out= subsetsWithDup(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

#include <iostream>#include <vector>#include <algorithm>#include <set>#include <iterator>using namespace std;//迭代；增量构造法//vector<vector<int> > subsetsWithDup(vector<int> &S) //{//sort(S.begin(), S.end()); // 必须排序//vector<vector<int> > result(1);//size_t previous_size = 0;//for (size_t i = 0; i < S.size(); ++i) //{//const size_t size = result.size();//for (size_t j = 0; j < size; ++j) //{//if (i == 0 || S[i] != S[i-1] || j >= previous_size)//{//result.push_back(result[j]);//result.back().push_back(S[i]);//}//}//previous_size = size;//}//return result;//}//迭代；二进制法vector<vector<int> > subsetsWithDup(vector<int> &S) {sort(S.begin(), S.end()); // 必须排序// 用set 去重，不能用unordered_set，因为输出要求有序set<vector<int> > result;const size_t n = S.size();vector<int> v;for (size_t i = 0; i < 1U << n; ++i) {for (size_t j = 0; j < n; ++j) {if (i & 1 << j)v.push_back(S[j]);}result.insert(v);v.clear();}vector<vector<int> > real_result;copy(result.begin(), result.end(), back_inserter(real_result));return real_result;}int main(){int a[3]={1,2,2};vector<int>vec(a,a+3);vector<vector<int>> out;out= subsetsWithDup(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: