[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] 

先想到的是背包...不过要得到所有结果，而不是结果数量，就用了DFS

写了一下，又参考了一下，有两种思路， 一种是统计各个数的需求个数，另一个是累加达到target 基本一致吧

class Solution {
private:
    vector<vector <int> > ret;
    vector <int> count;
public:
	void help(int current_index, int max_index, int target , vector<int> & candidates){

		if(target <0)
			return ;
        if(current_index == max_index)
		{
	    	if(target == 0)
			{
		        vector<int> tmp;
		    	for(int i = 0 ; i <  max_index; i++)
			    	for(int j = 0 ; j < count[i]; j++)
				    	tmp.push_back(candidates[i]);
		    	
		    	ret.push_back(tmp);
		    	return ;
	    	}
            return ;
         }
		for(int i = 0; i <= target/candidates[current_index] ; i++)
		{
			count[current_index] = i;
			help(current_index+1,max_index, target - i*candidates[current_index],candidates);
		}
	}
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
     ret.clear();
    if(candidates.size()==0)
        return ret;
    sort (candidates.begin(),candidates.end()); 
    count.clear();
    count.resize(candidates.size());
	help(0,candidates.size(), target,candidates);
	return  ret;
    }
};