題目：

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

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

Note:

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

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[
  [7],
  [2, 2, 3]
]

題解：

Solution 1 ()

class Solution {
public:
    void dfs(vector<vector<int>>& vv, vector<int>& v, vector<int> candidates, int target, int sum) {
        if
 
(sum == target) {
            vector<int>* tmp = new vector<int>;
            *tmp = v;
            sort((*tmp).begin(), (*tmp).end());
            if(find(vv.begin(), vv.end(), *tmp) == vv.end())
                vv.push_back(*tmp);
            delete tmp;
            return;
        }
        
 
if(sum > target) return;
        for(int i=0; i<candidates.size(); ++i) {
            v.push_back(candidates[i]);
            dfs(vv, v, candidates, target, sum+candidates[i]);
            v.pop_back();
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> vv;
        vector<int> v;
        dfs(vv, v, candidates, target, 0);
        return vv;        
    }
};

【LeetCode】039. Combination Sum