Subsets------DFS

最新推荐文章于 2023-05-29 14:46:26 发布

zjj2015

最新推荐文章于 2023-05-29 14:46:26 发布

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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算法。该算法确保子集中元素按非递减顺序排列且不含重复子集。通过递归辅助函数实现，每次迭代将当前元素加入现有子集中。

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

Have you met this question in a real interview? Yes
Example
If S = [1,2,3], a solution is:

[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]
Note
Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.

public class Solution {
    public ArrayList<ArrayList<Integer>> subsets(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(num == null || num.length == 0) {
            return result;
        }
        ArrayList<Integer> list = new ArrayList<Integer>();
        Arrays.sort(num);  
        subsetsHelper(result, list, num, 0);

        return result;
    }


    private void subsetsHelper(ArrayList<ArrayList<Integer>> result,
        ArrayList<Integer> list, int[] num, int pos) {

        result.add(new ArrayList<Integer>(list));

        for (int i = pos; i < num.length; i++) {

            list.add(num[i]);
            subsetsHelper(result, list, num, i + 1);
            list.remove(list.size() - 1);
        }
    }
}