SubSets，SubSets2, 求数组所有子集

问题描述：

Given a set of distinct integers, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

算法分析：第一种方法循环遍历方法，每次在原有集合基础上添加新元素。第二种算法：该类问题属于Backtraking回溯算法。

public class SubSets
{
	//循环遍历法
	public ArrayList<ArrayList<Integer>> subsets(int[] s)
	{
		if(s == null)
		{
			return null;
		}

		Arrays.sort(s);

		ArrayList<ArrayList<Integer>> res = new ArrayList<>();

		for(int i = 0; i < s.length; i ++)
		{
			ArrayList<ArrayList<Integer>> temp = new ArrayList<>();
			//get sets that are already in result
			for (ArrayList<Integer> a : res)
			{
				temp.add(new ArrayList<Integer>(a));
			}
			//add S[i] to existing sets
			for (ArrayList<Integer> a : temp)
			{
				a.add(s[i]);
			}
			//add S[i] only as a set
			ArrayList<Integer> single = new ArrayList<>();
			single.add(s[i]);
			temp.add(single);

			res.addAll(temp);
		}
		//add empty set
		res.add(new ArrayList<Integer>());

		return res;
	}

	//回溯法
	public List<List<Integer>> subsets_backtaking(int[] nums)
	{
		List<List<Integer>> res = new ArrayList<>();
		Arrays.sort(nums);
		dfs(res, new ArrayList<>(), nums, 0);
		return res;
	}
	public void dfs(List<List<Integer>> res, List<Integer> temp, int[] nums, int start)
	{
		res.add(new ArrayList<>(temp));
		for(int i = start; i < nums.length; i ++)
		{
			temp.add(nums[i]);
			dfs(res, temp, nums, i + 1);
			temp.remove(temp.size() - 1);
		}
	}
}

SubSets2：数组里有重复元素

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

Note: 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],
  []
]

public class SubSets2
{
	//回溯法
	public List<List<Integer>> subsets(int[] nums)
	{
		if(nums == null)
		{
			return null;
		}
		Arrays.sort(nums);//千万别忘了排序，否则就出错了，因为这个数组里面有重复元素
		List<List<Integer>> res = new ArrayList<>();
		dfs(res, new ArrayList<>(), nums, 0);
		return res;
	}
	public void dfs(List<List<Integer>> res, List<Integer> temp, int[] nums, int start)
	{
		res.add(new ArrayList<>(temp));
		for(int i = start; i < nums.length; i ++)
		{
			//skip duplicates
			if(i > start && nums[i] == nums[i-1])
			{
				continue;
			}
			temp.add(nums[i]);
			dfs(res, temp, nums, i + 1);
			temp.remove(temp.size() - 1);
		}
	}
}