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]

这道题看过好几次最开始都没看出怎么做就一直留着，结果今天在做Subsets 这道题时找到了灵感，使用深度优先的方法搜索，并且用一个vector记录向量，找到合适的向量时即将它保存在结果中，并进行回溯操作。这道题相比于Subsets中使用回溯法而言更麻烦一些。以后需要注意这种解题方法，当要求解的结果是一系列向量的集合时使用dfs搜索记录路径这种方法。对于输入candidates=[1,2] ，target=3，遍历的方向如图：

runtime:28ms<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4NCjxwcmUgY2xhc3M9"brush:java;"> class Solution { public: vector > combinationSum(vector & candidates, int target) { vector > result; vector path; sort(candidates.begin(),candidates.end()); helper(candidates,0,0,target,path,result); return result; } void helper(vector &nums,int pos,int base,int target,vector & path,vector > & result) { if(base==target) { result.push_back(path); return ; } if(base>target) return ; for(int i=pos;i

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