leetcode------Combination Sum II

标题：	Combination Sum II
通过率：	25.1%
难度：	中等

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]

与第一个版本不同的是set中的元素不能重复使用，那么递归时起始位置就要+1操作了代码如下：

 1 public class Solution {
 2     public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
 3         ArrayList<ArrayList<Integer>> res=new ArrayList<ArrayList<Integer>>();
 4         ArrayList<Integer> tmp=new ArrayList<Integer>();
 5         Arrays.sort(num);
 6         dfs(num,target,0,res,tmp);
 7         return res;
 8     }
 9     public void dfs(int[] num, int target,int start,ArrayList<ArrayList<Integer>> res,ArrayList<Integer> tmp){
10         if(target<0)return;
11         if(target==0 && !res.contains(tmp)){
12             res.add(new ArrayList<Integer>(tmp));
13             return ;
14         }
15         for(int i=start;i<num.length;i++){
16             tmp.add(num[i]);
17             dfs(num,target-num[i],i+1,res,tmp);
18             tmp.remove(tmp.size()-1);
19         }
20     }
21 }