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]
class Solution { public: std::vector<std::vector<int> > combinationSum(std::vector<int> &candidates, int target) { std::sort(candidates.begin(), candidates.end()); dfs(candidates, 0,0, target); return res; } private: std::vector<int> path; std::vector<std::vector<int>> res; // start 从candidates開始的位置, sum当前的和,target目标值,path存放满足条件的值,res 存放全部结果 void dfs(std::vector<int> &candidates, int start, int sum, int target) { if(sum>target)//超出目标值，直接返回 return ; if(sum == target)// 满足条件 { res.push_back(path); return ; } for(int i=start; i< candidates.size(); i++) { path.push_back(candidates[i]); dfs(candidates, i, sum+candidates[i], target); path.pop_back(); } } };
转载于:https://www.cnblogs.com/bhlsheji/p/4316664.html
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