Problem Description:

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] 

分析：按照题意要在给定数组中找出和为target的所有组合情况，其中每个元素都可以重复被选择多次，这是一个很典型的利用深度搜索寻找解的过程，将所有可能的情况都遍历出来，得到所有的组合，记录下来即可，代码如下：

class Solution {
public:
    void dfs(vector<int> &candidates,int target,int start,vector<vector<int> > &results,vector<int> &subset)
    {
        if(target==0)
        {
            results.push_back(subset);
            return;
        }
        if(target<0) 
            return;
        for(size_t i=start;i!=candidates.size();++i)
        {
            if(target<candidates[i]) return;
            subset.push_back(candidates[i]);
            dfs(candidates,target-candidates[i],i,results,subset);
            subset.pop_back();
        }
        
    }


    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int> > results;
        vector<int> subset;
        dfs(candidates,target,0,results,subset);
        return results;
    }
};