[Array]Combination Sum

Given a set of candidate numbers (C) 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]
]

方法１:简单的递归调用。

class Solution {
private:
    void recursive(int sum,vector<int> &combination,vector<int>& candidates,vector<vector<int>> &res,int target,int now){
            if(sum>target)
               return;
            if(sum==target){
              res.push_back(combination);
              return;
            }
            for(int i=now;i<candidates.size();i++){
                if(i!=now&&candidates[i]==candidates[i-1])
                    continue;
                combination.push_back(candidates[i]);
                recursive(sum+candidates[i],combination,candidates,res,target,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());
            for(int i=0;i<candidates.size();i++){
                if(i!=0&&candidates[i]==candidates[i-1])
                    continue;
                combination.push_back(candidates[i]);
                recursive(candidates[i],combination,candidates,res,target,i);
                combination.pop_back();
            }
            return res;
    }

};