解题思路
从 start 下标开始遍历候选数,每次可选当前数「可重复选,因此递归仍传当前下标 i」并扣减目标值,目标值为 0 时记录组合,回溯后尝试下一个数,避免重复组合
参考代码
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| class Solution { private List<List<Integer>> res = new ArrayList<>(); private List<Integer> tmp = new ArrayList<>();
public List<List<Integer>> combinationSum(int[] candidates, int target) { dfs(candidates, target, 0, res, tmp); return res; }
private void dfs(int[] candidates, int target, int start, List<List<Integer>> res, List<Integer> tmp) { if(target == 0) { res.add(new ArrayList<>(tmp)); return; } for(int i = start; i < candidates.length; i ++) { if(target - candidates[i] >= 0) { tmp.add(candidates[i]); dfs(candidates, target - candidates[i], i, res, tmp); tmp.remove(tmp.size() - 1); } } } }
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