Java for LeetCode 090 Subsets II

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],
  []
]

解题思路一：

偷懒做法，将Java for LeetCode 078 Subsets中的List换为Set即可通过测试，JAVA实现如下：

public List<List<Integer>> subsetsWithDup(int[] nums) {
	    Set<List<Integer>> list = new HashSet<List<Integer>>();
	    list.add(new ArrayList<Integer>());
	    Arrays.sort(nums);
	    for(int i=1;i<=nums.length;i++)
	        dfs(list, nums.length, i, 0,nums,-1);
	    return new ArrayList(list);
	}

	static List<Integer> alist = new ArrayList<Integer>();

	static void dfs(Set<List<Integer>> list, int n, int k, int depth,int[] nums,int last) {
	    if (depth >= k) {
	        list.add(new ArrayList<Integer>(alist));
	        return;
	    }
	    for (int i = last+1; i <= n-k+depth; i++) {
	        alist.add(nums[i]);
	        dfs(list, n, k, depth + 1,nums,i);
	        alist.remove(alist.size() - 1);
	    }
	}

解题思路二：

思路一其实用到了Java for LeetCode 077 Combinations和Java for LeetCode 078 Subsets的结论，使用set每次添加元素都需要查询一遍，会增加额外的时间开销，我们可以有一个不使用Set的解法，JAVA实现如下：

	public List<List<Integer>> subsetsWithDup(int[] S) {
		List<List<Integer>> res = new ArrayList<List<Integer>>();
		List<Integer> cur = new ArrayList<Integer>();
		Arrays.sort(S);
		dfs(0, res, cur, S);
		return res;
	}

	private void dfs(int dep, List<List<Integer>> res, List<Integer> cur,
			int[] S) {
		if (dep == S.length) {
			res.add(new ArrayList<Integer>(cur));
			return;
		}
		int upper = dep;
		while (upper >= 0 && upper < S.length - 1 && S[upper] == S[upper + 1])
			upper++;
		dfs(upper + 1, res, cur, S);
		for (int i = dep; i <= upper; i++) {
			cur.add(new Integer(S[dep]));
			dfs(upper + 1, res, cur, S);
		}
		for (int i = dep; i <= upper; i++)
			cur.remove(cur.size() - 1);
	}