Problem Description:
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], [] ]
分析:要求打印出所有不重复的组合,可以利用一个flag数组记录当前元素是否已被使用,然后判断之前该元素是否已被重复使用,所有组合为包含和不包含当前元素的总和,具体代码如下:
class Solution { public: void dfs(vector&s, int beg, int len, vector &mid, vector > &res,vector &flag) { if(len==0) { res.push_back(mid); return; } if(beg==s.size()) return; if(beg==0||s[beg]!=s[beg-1]||(s[beg]==s[beg-1]&&flag[beg-1]==1))//选中当前元素 { mid.push_back(s[beg]); flag[beg]=1; dfs(s,beg+1,len-1,mid,res,flag); flag[beg]=0; mid.pop_back(); dfs(s,beg+1,len,mid,res,flag); } else //不选中当前元素 { dfs(s,beg+1,len,mid,res,flag); } } vector > subsetsWithDup(vector &S) { vector > res; int len=S.size(); if(len==0) return res; vector mid; vector flag(len,0); sort(S.begin(),S.end()); for(int i=0;i<=len;++i)//长度为0到len的组合 dfs(S,0,i,mid,res,flag); } };