【LeetCode】Subsets I && II

1、Subsets

Total Accepted: 7967 Total Submissions: 29509 My Submissions

Given a set of distinct integers, S, 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 S = [1,2,3], a solution is:

[

[3],

[1],

[2],

[1,2,3],

[1,3],

[2,3],

[1,2],

[]

]

2、Subsets II

Total Accepted: 6207 Total Submissions: 23116 My Submissions

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

[

[2],

[1],

[1,2,2],

[2,2],

[1,2],

[]

]

求子序列集合，一个包含重复数据，一个不包含重复数据。

最后结果要求按照升序排列，并且根据例子来看，应该包含空集。

DFS，没什么可说的。BFS应该也可以解决，不过我没有写代码。有兴趣的可以试试。

这俩题的代码可以写成完全一样的，也是一个好的思维习惯。考虑这种问题的时候，一定要注意重复数据的处理。

这个题和Permutations &&Permutations
II，解题思路可以参考(LeetCode/Permutations && Permutations II)有点类似，本题比这个要更复杂一些。建议先思考简单的，再回来解决本题。

1、Java AC

public class Solution {
    public int len;
	public ArrayList<ArrayList<Integer>> subsets(int[] S) {
		ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
		if (S == null || S.length == 0) {
			return list;
		}
		len = S.length;
		Arrays.sort(S);
		for (int i = 1; i < len + 1; i++) {
			ArrayList<Integer> numList = new ArrayList<Integer>();
			dfs(list, numList, S, 0, i);
		}
		list.add(new ArrayList<Integer>());
		return list;
	}

	public void dfs(ArrayList<ArrayList<Integer>> list,
			ArrayList<Integer> numList, int[] S, int start, int k) {
		if (numList.size() == k) {
			list.add(new ArrayList<Integer>(numList));
			return;
		}
		ArrayList<Integer> allList = new ArrayList<Integer>();
		for (int i = start; i < len; i++) {
			if (allList.contains(S[i])) {
				continue;
			}
			allList.add(S[i]);
			numList.add(S[i]);
			dfs(list, numList, S, i + 1, k);
			numList.remove(numList.size() - 1);
		}
	}
}

2、Java AC

public class Solution {
    public int len;
	public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] num) {
		ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
		if (num == null || num.length == 0) {
			return list;
		}
		len = num.length;
		Arrays.sort(num);
		for (int i = 1; i < len + 1; i++) {
			ArrayList<Integer> numList = new ArrayList<Integer>();
			dfs(list, numList, num, 0, i);
		}
		list.add(new ArrayList<Integer>());
		return list;
	}
	public void dfs(ArrayList<ArrayList<Integer>> list,
			ArrayList<Integer> numList, int[] num, int start, int k) {
		if (numList.size() == k) {
			list.add(new ArrayList<Integer>(numList));
			return;
		}
		ArrayList<Integer> allList = new ArrayList<Integer>();
		for (int i = start; i < len; i++) {
			if (allList.contains(num[i])) {
				continue;
			}
			allList.add(num[i]);
			numList.add(num[i]);
			dfs(list, numList, num, i + 1, k);
			numList.remove(numList.size() - 1);
		}
	}
}