Brush title LeetCode notes - backtracking - the sum of a combination of problems

Subject description:

"Problems sum combination" Given a non-repeating array element candidates and a target number target, can find all the candidates for the target numbers and combinations thereof.

Unlimited numbers of candidates may be selected is repeated.

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/combination-sum

Java code implementation

    public static List<List<Integer>> combinationSum(int[] candidates, int target) {
　　　　　List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(candidates); //升序排序，便于剪枝
        combine2(result,new ArrayList<>(),candidates,target,0);
        return result;
    }


    public static void combine2(List<List<Integer>> result,List<Integer> temp,int[] candidates,int target,int begin){
        if (target==0){
            result.add(new ArrayList<>(temp));
            return;
        }

        for (int i=begin;i<candidates.length && target-candidates[i] >=0 ;i++){
            temp.add(candidates[i]);
            combine2(result,temp,candidates,target-candidates[i],i);
            temp.remove(temp.size()-1);
        }
    }