[LeetCode] Subsets II

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],
  []
]

用一个数组来记录某个数字是否被使用过，如果前面的数字和本数相同，则前面的数必须被使用过本数才能被使用。

 1 class Solution {
 2 private:
 3     vector<vector<int> > ret;
 4     bool canUse[100];
 5 public:
 6     void dfs(int dep, int maxDep, vector<int> &num, vector<int> a, int start)
 7     {
 8         ret.push_back(a);
 9         
10         if (dep == maxDep)
11             return;
12             
13         for(int i = start; i < num.size(); i++)
14             if (i == 0)
15             {
16                 canUse[i] = false;
17                 vector<int> b(a);
18                 b.push_back(num[i]);
19                 dfs(dep + 1, maxDep, num, b, i + 1);
20                 canUse[i] = true; 
21             }
22             else
23             {
24                 if (num[i] == num[i-1] && canUse[i-1])
25                     continue;
26                     
27                 canUse[i] = false;
28                 vector<int> b(a);
29                 b.push_back(num[i]);
30                 dfs(dep + 1, maxDep, num, b, i + 1);
31                 canUse[i] = true;
32             }          
33     }
34     
35     vector<vector<int> > subsetsWithDup(vector<int> &S) {
36         // Start typing your C/C++ solution below
37         // DO NOT write int main() function
38         sort(S.begin(), S.end());
39         ret.clear();
40         memset(canUse, true, sizeof(canUse));
41         vector<int> a;
42         dfs(0, S.size(), S, a, 0);
43         return ret;
44     }
45 };