subsetsII leetcode

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

public class Solution {
    public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(num.length == 0|| num == null){
            return result;
        }
        ArrayList<Integer> tem = new ArrayList<Integer>();
        Arrays.sort(num);
        dfs(result, tem, num, 0);
        return result;
    }
    public void dfs(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> tem, int[] num, int pos){
        result.add(new ArrayList<Integer> (tem));
        for(int i=pos; i<num.length; i++){
            if ( i != pos && num[i] == num[i - 1]) {
                continue;
            }    
            tem.add(num[i]);
            dfs(result,tem,num,i+1);
            tem.remove(tem.size()-1);
            
        }
    }
}

1. 这道题之前一定要sort一下。
2.判重用了一个if语句
分析：
假设还用subset方法 对于[1,2,2,2]
结果是[]
[1]
[1, 2(1)]
[1, 2(1), 2(2)]
[1, 2(1), 2(2), 2(3)]
[1, 2(1), 2(3)]//把2（3）删去 再把2(2)删去,此时2(1)层没循环完, 还可以再加入2(3)
[1,2(2)]
[1,2(2),(2,3)]
.............
只关心取了几个2不在乎取哪几个

在if语句里 i!= pos, 说明pos位置的数上一次已经取过(已跳过),为了避免与已经跳过的数重复, 说以如果num[i]==num[i-1]就continue