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].

 1 class Solution {
 2 public:
 3     vector<vector<int> > res;
 4     
 5     void permuteUnique(vector<int> &num, vector<int> &per,  vector<bool> &isAvail)
 6     {
 7         if(per.size() == num.size()) {
 8             res.push_back(per);
 9         }
10         int last_index = -1; // save arranged element
11         for(int i = 0; i < num.size(); i++) {
12             if(isAvail[i]) {
13                 if(last_index != -1 && num[i] == num[last]) continue;
14                 isAvail[i] = false;
15                 per.push_back(num[i]);
16                 permuteUnique(num, per, isAvail);
17                 per.pop_back();
18                 isAvail[i] = true;
19                 last_index = i;
20             }
21         }
22     }
23     
24     vector<vector<int> > permuteUnique(vector<int> &num) {
25         sort(num.begin(), num.end());
26         vector<int> per;
27         vector<bool> avail(num.size(), true);
28         permuteUnique(num, per, avail);
29         return res;
30     }
31 };