<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 fun( vector<vector<int>>&ans, vector<int>tmp, vector<int>&v,int t,int i){
        if(t==0){if(!tmp.empty())ans.push_back(tmp); return;}
        if(i==v.size())return;
        //take
        if(v[i]<=t){
            tmp.push_back(v[i]);
            fun(ans,tmp,v,t-v[i],i+1);
            tmp.pop_back();
            //not taking
            while(i<v.size()-1 && v[i]==v[i+1])i++;
            fun(ans,tmp,v,t,i+1);
        }

        return;

    }
    vector<vector<int>> combinationSum2(vector<int>& v, int t) {
        sort(v.begin(),v.end());
        vector<vector<int>>ans;
        vector<int>tmp;
        fun(ans,tmp,v,t,0);
        return ans;
    }
};
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