<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 solve(int i,int val,vector<int> a,vector<int> b){
    if(val==0){
        sort(b.begin(),b.end());
    ans.push_back(b);
    return;
    }
    else{
       for(int j=i;j<a.size();j++){
        if(j!=i && a[j]==a[j-1]) continue;
        if(val>=a[j]){
b.push_back(a[j]);
solve(j+1,val-a[j],a,b);
b.pop_back();
        }
       }
    return;
    }
}
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<int> b;
        sort(candidates.begin(),candidates.end());
        solve(0,target,candidates,b);
    return ans;
    }
};
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