Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

The same repeated number may be chosen from candidates unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

 

Constraints:

• 1 <= candidates.length <= 30
• 1 <= candidates[i] <= 200
• Each element of candidate is unique.
• 1 <= target <= 500

  Permutation: order matters [1, 2] and [2, 1] are different Combination: order does not matter [1, 2] and [2, 1] are the same 为了规避duplicates, 我们可以让结果里数字的排列顺序等于他们在初始数组里的排列顺序，也就是dfs到下一层的时候，只从当前num开始往后找。 dfs头里这个pos指的是我们从nums的哪个位置往后找。   实现：

class Solution {
private:
    void dfs(vector<int>& candidates, int target, vector<vector<int>>& res, vector<int>& combination, int pos){
        if (target == 0) res.push_back(combination);
        for (int i = pos; i < candidates.size(); i++){
            if (candidates[i] > target)
                return;
            combination.push_back(candidates[i]);
            dfs(candidates, target-candidates[i], res, combination, i);
            combination.pop_back();
        }
    }
    
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        
        vector<vector<int>> res;
        vector<int> combination;
        sort(candidates.begin(), candidates.end());
        dfs(candidates, target, res, combination, 0);
        return res;
        
    }
};