<p>Given a collection of candidate numbers (<code>candidates</code>) and a target number (<code>target</code>), find all unique combinations in <code>candidates</code>&nbsp;where the candidate numbers sum to <code>target</code>.</p>

<p>Each number in <code>candidates</code>&nbsp;may only be used <strong>once</strong> in the combination.</p>

<p><strong>Note:</strong>&nbsp;The solution set must not contain duplicate combinations.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> candidates = [10,1,2,7,6,1,5], target = 8
<strong>Output:</strong> 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> candidates = [2,5,2,1,2], target = 5
<strong>Output:</strong> 
[
[1,2,2],
[5]
]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;=&nbsp;candidates.length &lt;= 100</code></li>
	<li><code>1 &lt;=&nbsp;candidates[i] &lt;= 50</code></li>
	<li><code>1 &lt;= target &lt;= 30</code></li>
</ul>

class Solution {
public:
void help(int i,int k,vector<int> temp,vector<int> v,set<vector<int>> &res){
        if(k==0){
            sort(temp.begin(),temp.end());
            res.insert(temp);
            return;
        }
        for(int j=i; j<v.size(); j++){
            if(j>i&&v[j]==v[j-1]) continue;
            if(v[j]>k) break;
            temp.push_back(v[j]);
            help(j+1, k-v[j], temp, v, res);
            temp.pop_back();
        }
    }
    vector<vector<int>> combinationSum2(vector<int>& v, int k) {
        set<vector<int>> res;
        vector<int> temp;
        sort(v.begin(),v.end());
        help(0,k,temp,v,res);
        vector<vector<int>> ans;
        ans.assign(res.begin(),res.end());
        return ans;
    }
};
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