LeetCode 90. Subsets II

原题链接在这里：https://leetcode.com/problems/subsets-ii/

题目：

Given a collection of integers that might contain duplicates, nums, 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 nums = [1,2,2], a solution is:

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

题解：

是Subsets的进阶版. 这里有duplicates, e.g. [1,2,2]但是res中不能包含两个[2].

还是backtracking, 注意去重.

Time Complexity: exponential.

Space: O(nums.length). stack space.

AC Java:

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

也所以在elem加完新元素想要放回res之前，需要先判断res中是否含有这个elem, 若是没有可以加到res中，若是已经有了，就不可以加到res中.

Time Complexity: exponential.

Space: O(res.size()).

AC Java:

 1 public class Solution {
 2     public List<List<Integer>> subsetsWithDup(int[] nums) {
 3         List<List<Integer>> res = new ArrayList<List<Integer>>();
 4         if(nums == null || nums.length == 0){
 5             return res;
 6         }
 7         Arrays.sort(nums);
 8         res.add(new ArrayList<Integer>());
 9         for(int i = 0; i<nums.length; i++){
10             int size = res.size();
11             for(int j = 0; j<size; j++){
12                 ArrayList<Integer> elem = new ArrayList<Integer>(res.get(j));
13                 elem.add(nums[i]);
14                 if(!res.contains(elem)){
15                     res.add(elem);
16                 }
17             }
18         }
19         return res;
20     }
21 }