39. 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.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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]

求vector中哪些数字之和为target，可以有重复的。

class Solution {
public:
    vector<vector<int>> ans;
    void dfs(vector<int> c,int index,int sum,int target,vector<int> &path){
        if(sum>target)return;
        else if(sum==target){
            ans.push_back(path);
            return;

        }
        for(int i= index; i<c.size();i++){
            path.push_back(c[i]);
            dfs(c,i,sum+c[i],target,path);
            path.pop_back();
        }
    }
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<int> path;
        dfs(candidates,0,0,target,path);
        return ans;
    }
};