Combination Sum

Q: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] 

A: DFS问题。

注意迭代结束条件：target==0，发现解; target<0，无解

 

   vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > result;
        vector<int> path;
        
        sort(candidates.begin(),candidates.end());
        combinationsum_aux(0,candidates,target,path,result);
        
        return result;
        
    }
    
    bool combinationsum_aux(int curPos,vector<int>& candidates,int target,vector<int>& path,vector<vector<int> >& result)
    {
        if(target==0)
        {
            result.push_back(path);
            return true;
        }else if(target<0)
            return false;
        
        for(int i=curPos;i<candidates.size();i++)
        {
            path.push_back(candidates[i]);
            if(!combinationsum_aux(i,candidates,target-candidates[i],path,result))  //注意i不是加一，因为数可以无限次被使用。这里还做了剪枝，当返回false，说明当前层（）已经完成了搜索，返回到上层。
            {
                path.pop_back();
                break;
            }
            path.pop_back();
        }
        return true;
    }