<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:
    vector<vector<int>> ans;
    void dfs(int now, vector<int>& sans, int target, vector<int>& candidates, int pl){
        if(now==target){
            ans.push_back(sans);
            return;
        }
        if(now>target || pl>=candidates.size()) return;
        int tot = 1;
        int cur = candidates[pl];
        int old_pl = pl;
        pl++;
        sans.push_back(cur);
        while(pl<candidates.size()&&candidates[pl]==cur) pl++, tot++, sans.push_back(cur);
        dfs(now + tot*cur, sans, target, candidates, pl);
        for(int i=0;i<tot;i++) sans.pop_back();
        dfs(now, sans, target, candidates, old_pl + 1);
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<int>tmp;
        dfs(0, tmp, target, candidates, 0);
        return ans;
    }
};
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