78, Subsets

I Problem

Given an integer array nums of unique elements, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

Example 1
Input: nums = [1, 2, 3]
Output: [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

Example 2
Input: nums = [0]
Output: [[], [0]]

Constraints

• 1 <= nums.length <= 10
• -10 <= nums[i] <= 10
• All the numbers of nums are unique.

Related Topics

• Array
• Backtracking
• Bit Manipulation

II Solution

Approach 1: Backtracking

pub fn subsets(nums: Vec<i32>) -> Vec<Vec<i32>> {
    const DFS: fn(usize, &[i32], &mut Vec<i32>, &mut Vec<Vec<i32>>) = 
        |i, nums, subset, res| {
            res.push(subset.clone());

            if i == nums.len() {
                return;
            }

            for j in i..nums.len() {
                subset.push(nums[j]);
                DFS(j + 1, nums, subset, res);
                subset.pop();
            }
        };
    let mut res = Vec::with_capacity(2_usize.pow(nums.len() as u32));

    DFS(0, &nums, &mut vec![], &mut res);

    res
}

@FunctionalInterface
interface QuadrConsumer<A, B, C, D> {
    void accept(A a, B b, C c, D d);
}

QuadrConsumer<Integer, int[], List<Integer>, List<List<Integer>>> dfs =
        (i, nums, subset, res) -> {
            res.add(new ArrayList<>(subset));

            if (i == nums.length) {
                return;
            }

            for (int j = i; j < nums.length; j++) {
                subset.addLast(nums[j]);
                this.dfs.accept(j + 1, nums, subset, res);
                subset.removeLast();
            }
        };

public List<List<Integer>> subsets(int[] nums) {
    List<List<Integer>> res = new ArrayList<>();
    this.dfs.accept(0, nums, new ArrayList<>(), res);
    return res;
}

Approach 2: Iteration

pub fn subsets(nums: Vec<i32>) -> Vec<Vec<i32>> {
    let mut res = vec![vec![]];

    for i in 0..nums.len() {
        for j in 0..res.len() {
            let mut subset = res[j].clone();

            subset.push(nums[i]);

            res.push(subset);
        }
    }

    res
}

public List<List<Integer>> subsets(int[] nums) {
    List<List<Integer>> res = new ArrayList<>() {{
        this.add(new ArrayList<>());
    }};

    for (int i = 0; i < nums.length; i++) {
        for (int j = 0, size = res.size(); j < size; j++) {
            ArrayList<Integer> subset = new ArrayList<>(res.get(j));

            subset.add(nums[i]);
            
            res.add(subset);
        }
    }

    return res;
}