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:

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        dfs(nums, 0, new ArrayList<Integer>(), res);
        return res;
    }
    
    private void dfs(int [] nums, int start, List<Integer> item, List<List<Integer>> res){
        res.add(new ArrayList<Integer>(item));
        for(int i = start; i<nums.length; i++){
            if(i>start && nums[i]==nums[i-1]){
                continue;
            }
            item.add(nums[i]);
            dfs(nums, i+1, item, res);
            item.remove(item.size()-1);
        }
    }
}

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:

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(nums == null || nums.length == 0){
            return res;
        }
        Arrays.sort(nums);
        res.add(new ArrayList<Integer>());
        for(int i = 0; i<nums.length; i++){
            int size = res.size();
            for(int j = 0; j<size; j++){
                ArrayList<Integer> elem = new ArrayList<Integer>(res.get(j));
                elem.add(nums[i]);
                if(!res.contains(elem)){
                    res.add(elem);
                }
            }
        }
        return res;
    }
}