[leetcode] Permutations II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,

[1,1,2]

 have
the following unique permutations:

[1,1,2]

, 

[1,2,1]

,
and 

[2,1,1]

.
思路：先对num排序，然后递归，跳过重复元素

代码：

class Solution {
private:
int n;
vector<bool> flag;
vector<vector<int> > res;
public:
vector<vector<int> > permuteUnique(vector<int> &num) {
n=num.size();
res.clear();
flag.resize(n,false);
if(n==0) return res;
vector<int> temp(n,0);
sort(num.begin(),num.end());
dfs(temp,num,0);
return res;
}
void dfs(vector<int> temp, vector<int> &num, int step){
if(step==n){
res.push_back(temp);
return;
}
for(int i=0;i<n;i++){
if(flag[i]==false){
temp[step]=num[i];
flag[i]=true;
dfs(temp,num,step+1);
flag[i]=false;
while(i<n-1 && num[i+1]==num[i]) i++;
}
}
}
};