<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:
    set<vector<int>> S;
    vector<int> c;
    void getAll(vector<int> &candidates, int target, int i) {
        if(target == 0) {
            S.insert(c);
            return;
        }
        if(i == candidates.size() || target < 0) {
            return;
        }
        for(int k=i;k<candidates.size();k++) {
            if(k == i || candidates[k] != candidates[k-1]) {
                c.push_back(candidates[k]);
                getAll(candidates, target - candidates[k], k + 1);
                c.pop_back();
            }
        }
        // getAll(candidates, target, i + 1);
        // c.push_back(candidates[i]);
        // getAll(candidates, target-candidates[i], i + 1);
        // c.pop_back();
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        sort(candidates.begin(), candidates.end());
        getAll(candidates, target, 0);
        for(auto i=S.begin();i!=S.end();i++) {
            res.push_back(*i);
        }
        return res;
    }
};
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