<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 combination(vector<vector<int>>& ans, vector<int>& ds,vector<int> nums,int i,int s,int n, int target) {
        if(s == target) {
            ans.push_back(ds);
        }
        for(int k = i;k < n;k++) {
            if( k != i && nums[k] == nums[k-1]) continue;
            if(s+nums[i] > target) break;
            s += nums[k];
            ds.push_back(nums[k]);
            combination(ans,ds,nums,k+1,s,n,target);
            s -= nums[k];
            ds.pop_back();
        }
    }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        vector<vector<int>> ans;
        vector <int> ds;
        int n = candidates.size();
        sort(candidates.begin(),candidates.end());
        combination(ans, ds, candidates, 0, 0, n, target);
        return ans;
    }
};
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