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] 

给定一组候选解集合，以及一个目标值，给出所有能够累加得到该目标值的组合，候选解元素可以重复使用任意次。

题目要求解的形式要按非降序，那就开始之前将候选序列排个序。

 Solution {
public:
    /*
      cur_sum: 当前的累加和
      icur   : 当前候选数的下标
      target ：目标值
      cur_re : 当前的潜在解
      re     : 最终的结果集
      can    : 候选数集
    */
    void dfs(int cur_sum, int icur, int target, vector<int>& cur_re, vector<vector<int> > &re, vector<int> can)
    {
        if(cur_sum > target || icur >= can.size()) return;
        if(cur_sum == target){
            //cur_re.push_back(cur_sum);
            re.push_back(cur_re);
            return;
        }
        //(target - cur_sum >= can[i]类似剪枝，加上该判定与
        //不加，时间差4-5倍。
        for(int i = icur; i < can.size() && (target - cur_sum >= can[i]); ++i){
            cur_sum += can[i];
            cur_re.push_back(can[i]);
            dfs(cur_sum, i, target, cur_re, re, can);
            //go back
            cur_re.pop_back();
            cur_sum -= can[i];
        }
    }
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<vector<int> > re;
        vector<int> cur_re;
        if(candidates.size() == 0) return re;
        sort(candidates.begin(), candidates.end());
        if(candidates[0] > target) return re;
        dfs(0, 0, target, cur_re, re, candidates);
        return re;
    }
};

Combination Sum