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] 

思路：在集合中找出组合之和为目标值的集合。这让我们很容易联想到01背包问题，递归回溯累加到目标值。

class Solution {

public:

    void resolve(vector<int> &candidates,int target,int start,int sum,vector<int> &path,vector<vector<int> > &result)

    {

        if(sum>target)

            return;

        if(sum==target)

        {

            result.push_back(path);

            return;

        }

        for(int i=start;i<candidates.size();i++)

        {

            path.push_back(candidates[i]);

            resolve(candidates,target,i,sum+candidates[i],path,result);

            path.pop_back();

        }

    }

    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {

        vector<vector<int> > result;

        vector<int> path;

        sort(candidates.begin(),candidates.end());

        result.clear();

        path.clear();

        resolve(candidates,target,0,0,path,result);

        return result;

    }

};