<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;
    set<vector<int>>st;
    void rec(vector<int>&c, int t, int i, vector<int>&a,int pre){
        if(t==0){
            if(st.find(a)==st.end()){
                ans.push_back(a);
                st.insert(a);
            }
            return;
        }
        if(i>=c.size())
            return;
        if(t<0)
            return;
        rec(c,t,i+1,a,t);
        if(i>0 && c[i]==c[i-1] && pre==t) return;
        a.push_back(c[i]);
        rec(c,t-c[i],i+1,a,t);
        a.pop_back();
        
    }
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<int>v;
        sort(candidates.begin(),candidates.end());
        rec(candidates,target, 0, v,target);
        return ans;
    }
};
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