Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Input: [1,2,2]
Output:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

題意：

是的[leetcode]78. Subsets數組子集 follow up

這題強調給定數組可以有重復元素

思路：

需要先對給定數組排序，以便去重

代碼：

 1 class Solution {
 2     public List<List<Integer>> subsetsWithDup(int[] nums) {
 3         List<List<Integer>> result = new ArrayList<>();
 4         if(nums == null || nums.length ==0 )return result;
 5         Arrays.sort(nums);
 6         List<Integer> path = new
 
 ArrayList<>();
 7         dfs(0, nums, path, result);
 8         return result;    
 9     }
10     
11     private void dfs(int index, int[] nums, List<Integer> path, List<List<Integer>> result){
12         result.add(new ArrayList<>(path));
13         for(int i = index; i < nums.length; i++){
 
14             if( i != index && nums[i] == nums[i-1] ) continue;
15             path.add(nums[i]);
16             dfs(i + 1, nums, path, result);
17             path.remove(path.size() - 1);
18         }
19     }
20 }

[leetcode]90. Subsets II數組子集(有重)