Combination Sum (LeetCode HashMap)

Question:  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]

Idea: Depth-first search. First sort the array. Then for each of the numbers, e.g. num[i], smaller than the target, add num[i] to the path and recursively find the next target=target-num[i]. Since each element can be used unlimited number of times, the next start index is the same as the current i. If the next target==0, means this is a valid path (a solution), add it to the result.

Time: O(n^n) (actually O(n!)) Space: O(n)

Code:

 public class Solution {  
   public List<List<Integer>> combinationSum(int[] candidates, int target) {  
     List<List<Integer>> result=new ArrayList<List<Integer>>();  
     if(candidates.length==0)  
       return result;  
     Arrays.sort(candidates);  
     ArrayList<Integer> path=new ArrayList<Integer>();  
     dfs(candidates,target,0,path,result);  
     return result;  
   }  
   public void dfs(int[] nums,int target,int start,ArrayList<Integer> path,List<List<Integer>>result)  
   {  
     if(target==0)  
     {  
       result.add(new ArrayList<Integer>(path));  
       return;  
     }  
     for(int i=start;i<nums.length;i++)  
     {  
       if(nums[i]>target)  
         break;  
       if(i>start&&nums[i]==nums[i-1])  
         continue;  
       int nextTarget=target-nums[i];  
       path.add(nums[i]);  
       dfs(nums,nextTarget,i,path,result);  
       path.remove(path.size()-1);  
     }  
   }  
 }