[LeetCode]Subsets

最新推荐文章于 2019-04-17 12:01:23 发布

原创最新推荐文章于 2019-04-17 12:01:23 发布 · 560 阅读

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本文介绍了一种生成给定整数集合所有可能子集的方法。通过递归和增量构造法实现，确保子集按非递减顺序排列且不含重复。以[1,2,3]为例，展示了所有可能的子集。

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],
  []
]

递归

public class Solution {
	public List<List<Integer>> subsets(int[] S) {
	    Arrays.sort(S);
	    return help(S);
	}
	
	private List<List<Integer>> help(int[] S){
		if (S.length == 0) {
			List<List<Integer>> lin = new ArrayList<List<Integer>>();
			lin.add(new ArrayList<Integer>());
			return lin;
		}
		int a[] = Arrays.copyOf(S, S.length - 1);
		List<List<Integer>> lin = help(a);
		List<List<Integer>> res = new ArrayList<>(lin);
		for (int i = 0; i < lin.size(); i++) {
			List<Integer> list = new ArrayList<>(lin.get(i));
			list.add(S[S.length - 1]);
			res.add(list);
		}
		return res;
	}
}

增量构造法+递归

public class Solution {
	List<List<Integer>> res = new ArrayList<List<Integer>>();
	List<Integer> list = new ArrayList<Integer>();

	public List<List<Integer>> subsets(int[] s) {
		Arrays.sort(s);
		help(s, 0);
		return res;
	}

	private void help(int[] num, int start) {
		List<Integer> lin = new ArrayList<Integer>(list);
		res.add(lin);
		for (int i = start; i < num.length; i++) {
			list.add(num[i]);
			help(num, i + 1);
			list.remove(Integer.valueOf((num[i])));
		}
	}
}