Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
- All numbers (including
target) will be positive integers. - The solution set must not contain duplicate combinations.
Example 1:
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
1 class Solution { 2 public: 3 void getsum(vector<vector<int>>&res,vector<int> ans,vector<int>&candidates, int target, int lastidx) { 4 if (target == 0) { 5 res.push_back(ans); 6 return; 7 } 8 if (target < candidates[lastidx])return; 9 for (int i = lastidx; i < candidates.size(); i++) { 10 if (target < candidates[i])break; 11 ans.push_back(candidates[i]); 12 getsum(res, ans, candidates, target - candidates[i], i); 13 ans.pop_back(); 14 } 15 } 16 vector<vector<int>> combinationSum(vector<int>& candidates, int target) { 17 sort(candidates.begin(), candidates.end()); 18 vector<vector<int>>res; 19 vector<int>ans; 20 getsum(res, ans, candidates, target, 0); 21 return res; 22 } 23 };
感觉可以用dp，不知道会不会快