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, … , 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] 

class Solution {
/*思路，采用DFS，主要涉及sub()函数参数的设置。分支剪枝。。。调用DFS模板就可*/
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> re;
        vector<int> are;
        if(candidates.size()==0) return re;
        sub(re,candidates,are,target,0);
        return re;
    }
    void sub(vector<vector<int>> &result,vector<int> &candidates,vector<int> &aresult,int left,int level){
        if(left==0) result.push_back(aresult); 
        if(left<0) return; //剪枝
        for(int i=level;i<candidates.size();i++){  //需要注意的是，我只要将candidates[i]的取舍情况都列出来就可了。不要再想下一个元素了
            aresult.push_back(candidates[i]);
            sub(result,candidates,aresult,left-candidates[i],i); //考虑到当前元素
            aresult.pop_back(); //不考虑当前元素。对下一个元素在下一个循环中会讨论，不用再sub()调用了。
        }
    }
};