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] ]
Time: O(N^|target|)
Space: O(|target|)
1 class Solution: 2 def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]: 3 res = [] 4 lst = [] 5 self.dfs(candidates, lst, res, 0, target) 6 return res 7 8 def dfs(self, candidates, lst, res, start, reminder): 9 if reminder < 0: 10 return 11 if reminder == 0: 12 res.append(list(lst)) 13 return 14 for i in range(start, len(candidates)): 15 cur_num = candidates[i] 16 lst.append(cur_num) 17 self.dfs(candidates, lst, res, i, reminder - cur_num) 18 lst.pop() 19 20
class Solution { public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> result = new ArrayList<>(); if (candidates == null) { return result; } List<Integer> list = new ArrayList<>(); helper(candidates, list, 0, target, result); return result; } private void helper(int[] nums, List<Integer> list, int index, int reminder, List<List<Integer>> result) { if (reminder < 0) { return; } if (reminder == 0) { result.add(new ArrayList<>(list)); return; } for (int i = index; i < nums.length; i++) { list.add(nums[i]); helper(nums, list, i, reminder - nums[i], result); list.remove(list.size() - 1); } } }