Combination Sum II (LeetCode Array)

Question:  Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
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 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

Idea: Depth-first search. First sort the array. Then for each number smaller than the target, e.g. num[i], add it to the path and recursively search for the next target=target - num[i]. Since each number can be used only once, the start index of the recursion is i+1. If the current target==0, means a valid solution is fond. Add it to the result.

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

Code:

 public class Solution {  
   public List<List<Integer>> combinationSum2(int[] num, int target) {  
     List<List<Integer>> result=new ArrayList<List<Integer>>();  
     if(num==null||num.length==0)  
       return result;  
     Arrays.sort(num);  
     ArrayList<Integer> path=new ArrayList<Integer>();  
     dfs(num,target,0,path,result);  
     return result;  
   }  
   public void dfs(int[] num,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<num.length;i++)  
     {  
       if(num[i]>target)  
         break;  
       if(i>start&&num[i]==num[i-1])  
         continue;  
       int nextTarget=target-num[i];  
       path.add(num[i]);  
       dfs(num,nextTarget,i+1,path,result);  
       path.remove(path.size()-1);  
     }  
   }  
 }