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] 

题意分析：从给定数组中找到一组数字，要求这组数字之和等于target。另外，数组中的数字不允许被使用多次，但如果一开始就存在多个的话，可以使用多次。
解题思路：显然先排序，然后dfs。其中有一点要注意的是：因为不能重复，所以要跳过一样的数字。以上面为例，如果不跳过重复的1的话，会出现多个：[1,7]。

上AC代码：

public class Solution {        void dfs(int [] num, int start, int target, ArrayList<Integer> array, ArrayList<ArrayList<Integer>> result) {        if(target==0) {            result.add(new ArrayList<Integer>(array));            return;        }                if(start>=num.length||num[0]>target) {            return;        }        int i = start;        while(i<num.length) {            if(num[i]<=target) {                array.add(num[i]);                dfs(num, i + 1, target - num[i], array, result);                array.remove(array.size()-1);                //跳过重复的元素                while(i<(num.length-1)&&num[i]==num[i+1]) {                    i++;                }            }            i++;        }    }            public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();         ArrayList<Integer> array = new ArrayList<Integer>();         if(num==null) {             result.add(array);             return result;         }         Arrays.sort(num);         dfs(num,0, target,array,result);          return result;    }}