Leetcode 78. Subsets

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本文介绍了一种使用递归和位运算解决数组子集问题的方法，包括如何确保子集按非降序排列且不包含重复子集。

Question

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

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

Code

递归

public List<List<Integer>> subsets1(int[] S) {
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        if (S == null) {
            return ret;
        }

        Arrays.sort(S);

        dfs(S, 0, new ArrayList<Integer>(), ret);

        return ret;
    }

    public void dfs(int[] S, int index, List<Integer> path, List<List<Integer>> ret) {
        ret.add(new ArrayList<Integer>(path));

        for (int i = index; i < S.length; i++) {
            path.add(S[i]);
            dfs(S, i + 1, path, ret);
            path.remove(path.size() - 1);
        }
    }

位运算方式

 public List<List<Integer>> subsets(int[] nums) {

        Arrays.sort(nums);
        List<List<Integer>> results = new ArrayList<>();
        int len = (int) Math.pow(2, nums.length);
        boolean[] flags = new boolean[nums.length];
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < flags.length; j++) {
                if (!flags[j]) {
                    flags[j] = true;
                    break;
                }
                flags[j] = false;
            }
            List<Integer> sub = new ArrayList<>();
            for (int j = 0; j < flags.length; j++) {
                if (flags[j]) {
                    sub.add(nums[j]);
                }
            }
            results.add(sub);
        }
        return results;
    }