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]
Solution:
public List<List<Integer>> combinationSum(int[] candidates, int target) { List<List<Integer>> result = new ArrayList<>(); Arrays.sort(candidates); combine(result, new ArrayList<Integer>(), candidates, target, 0); return result; } private void combine(List<List<Integer>> result, List<Integer> list, int[] num, int target, int pos) { if(target < 0) return; if(target == 0) { result.add(new ArrayList<Integer>(list)); return; } for(int i=pos; i<num.length; i++) { list.add(num[i]); combine(result, list, num, target-num[i], i); //注意,从i开始,不是i+1,因为可以重复使用 list.remove(list.size()-1); } }