LeetCode - 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, ..., a_k) must be in non-descending order. (ie, a₁ <= a₂ <=...<= a_k).
• 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] 

class Solution {public:    void findPath(set<vector<int> >& numSet, vector<vector<int> >& dp,vector<int> &candidates, vector<int>& path,int value){if(value==0){vector<int> tmp(path);sort(tmp.begin(),tmp.end());numSet.insert(tmp);return;}for(int i=0;i<dp.size();i++){if(dp[i][value]!=-1){path.push_back(candidates[i]);findPath(numSet,dp,candidates,path,dp[i][value]);path.pop_back();}}    }     vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        vector<vector<int> > result;if(candidates.empty()||candidates.size()==0){return result;}int n=candidates.size();  sort(candidates.begin(),candidates.end());vector<vector<int> > dp(n,vector<int>(target+1,-1));set<vector<int> > numSet;vector<bool> visited(target+1,false);vector<int> path;visited[0]=true; for(int i=0;i<n;i++){if(candidates[i]>target){break;}for(int j=candidates[i];j<=target;j++){if(visited[j-candidates[i]]==true){dp[i][j]=j-candidates[i];visited[j]=true;}}}findPath(numSet,dp,candidates,path,target);result.assign(numSet.begin(),numSet.end());        return result;    }};