Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:
For example,
If S = [1,2,2], a solution
is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
给定一个可能包含重复值的整数集合S,返回所有可能的子集合
class Solution {
public:
void getCombinations(vector<vector<int> >&result, vector<int>&s, vector<int>combination, int index2add, int kth, int k){
//将s中的第index2add位上的数作为组合中的第kth个数
combination.push_back(s[index2add]);
if(kth==k){
result.push_back(combination);
return;
}
for(int i=index2add+1; i+(k-kth-1)<s.size(); i++){
if(i!=index2add+1 && s[i]==s[i-1])continue; //去重
getCombinations(result, s, combination, i, kth+1, k);
}
}
vector<vector<int> > getSubsetK(vector<int>&s, int k){
// 求长度为k的子集
vector<vector<int> >result;
vector<int> combination;
for(int i=0; i+(k-1)<s.size(); i++){
if(i!=0 && s[i]==s[i-1])continue; //排重
getCombinations(result, s, combination, i, 1, k);
}
return result;
}
vector<vector<int> > subsetsWithDup(vector<int> &S) {
vector<vector<int> >result;
vector<int> emptySet; //一定有个空集
result.push_back(emptySet);
int size=S.size();
if(size==0)return result;
sort(S.begin(), S.end());
for(int k=1; k<=size; k++){
//分别求出长度为1,2,3,...size的子集
vector<vector<int> >subsets=getSubsetK(S, k);
result.insert(result.end(), subsets.begin(), subsets.end());
}
return result;
}
};LeetCode: Subsets II [091],布布扣,bubuko.com
原文地址:http://blog.csdn.net/harryhuang1990/article/details/27821221