[Leetcode] Combination Sum II

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]

Solution:

 1 public class Solution {
 2     public List<List<Integer>> combinationSum2(int[] candidates, int target) {
 3         List<List<Integer>> result=new ArrayList<List<Integer>>();
 4         List<Integer> al=new ArrayList<Integer>();
 5         Arrays.sort(candidates);
 6         dfs(result,al,target,candidates,0);
 7         return result;
 8     }
 9 
10     private void dfs(List<List<Integer>> result, List<Integer> al, int target, int[] candidates, int position) {
11         // TODO Auto-generated method stub
12         if(target<0)
13             return;
14         if(target==0){
15             result.add(new ArrayList<Integer>(al));
16             return;
17         }
18         for(int i=position;i<candidates.length;++i){
19             al.add(candidates[i]);
20             dfs(result, al, target-candidates[i], candidates, i+1);
21             al.remove(al.size()-1);
22             while((i+1<candidates.length)&&candidates[i]==candidates[i+1])
23                 ++i;
24         }    
25     }
26 }

这道题和Combination Sum那道题很像，区别就是在第20行和第22行，

如果candidates[i]==candidates[i+1]，就直接跳过candidates[i+1]这个选项了，因为不能有重复的solution嘛。