Combination Sum

This question is typically a DP one. It might be the simplest one among those DP questions I have solved.

Basic idea is to start from the largest element in candidates array, then try to use this one as many times  as possible to let remainNum get 0. If remainNum gets 0, then remove latest added one, and lets try next number in array, until all numbers are traversed; otherwise, add new number into combination,

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
   
        ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
        Arrays.sort(candidates);
        if((target <= 0) || (candidates.length == 0) || (target < candidates[0])) return res;
        ArrayList<Integer> comb = new ArrayList<Integer>();
        findCombOfSums(res,comb,candidates, candidates.length - 1, target);
        return res;
    }
   
    public void findCombOfSums(ArrayList<List<Integer>> res, ArrayList<Integer> comb,int[] candidates, int index, int remainNum){
        if(index >= 0 && remainNum >= 0){
            if(remainNum == 0){
               
                ArrayList<Integer> oneRes = new ArrayList<Integer>();
                for(int i = comb.size()-1; i >= 0;i--){// reverse the order
                    oneRes.add(comb.get(i));
                }
                if(!res.contains(oneRes)) res.add(oneRes);

            }
            else{
                    comb.add(candidates[index]);
                    findCombOfSums(res,comb,candidates, index, remainNum - candidates[index]);
                    comb.remove(comb.size()-1);
                    findCombOfSums(res,comb,candidates, index - 1, remainNum);
            }
        }
    }
}