[leedcode 39] Combination Sum

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]  

public class Solution {
    List<Integer> seq;
    List<List<Integer>> res;
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        //        ，       ，for     ，       sum>target    ，  sum=target     
        //  for       ，       ，           。level       candidates  
        //    ：： res    ，   new  Integer，       ，     seq         ！
        Arrays.sort(candidates);
        seq=new ArrayList<Integer>();
        res=new ArrayList<List<Integer>>();
        find(candidates,target,0,0);

        return res;
    }
    public void find(int[] candidates,int target,int sum,int level){
        if(sum==target){
            res.add(new ArrayList<Integer>(seq));
            return;
        }
        if(sum>target){
            return;
        }
        for(int i=level;i<candidates.length;i++){
            seq.add(candidates[i]);
            sum+=candidates[i];
            find(candidates,target,sum,i);
            seq.remove(seq.size()-1);//
            sum-=candidates[i];
        }
        
    }
}