README_EN.md

78. Subsets

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Description

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.

Solutions

DFS.

Python3

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        res = []

        def dfs(i, n, t):
            res.append(t.copy())
            if i == n:
                return
            for j in range(i, n):
                t.append(nums[j])
                dfs(j + 1, n, t)
                t.pop()

        dfs(0, len(nums), [])
        return res

Java

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

    private void dfs(int i, int[] nums, List<Integer> t, List<List<Integer>> res) {
        res.add(new ArrayList<>(t));
        if (i == nums.length) {
            return;
        }
        for (int j = i; j < nums.length; ++j) {
            t.add(nums[j]);
            dfs(j + 1, nums, t, res);
            t.remove(t.size() - 1);
        }
    }
}

C++

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> t;
        dfs(0, nums, t, res);
        return res;
    }

    void dfs(int i, vector<int>& nums, vector<int> t, vector<vector<int>>& res) {
        res.push_back(t);
        if (i == nums.size()) return;
        for (int j = i; j < nums.size(); ++j)
        {
            t.push_back(nums[j]);
            dfs(j + 1, nums, t, res);
            t.pop_back();
        }
    }
};

Go

func subsets(nums []int) [][]int {
	var res [][]int
	var t []int
	dfs(0, nums, t, &res)
	return res
}

func dfs(i int, nums, t []int, res *[][]int) {
	cp := make([]int, len(t))
	copy(cp, t)
	*res = append(*res, cp)
	if i == len(nums) {
		return
	}
	for j := i; j < len(nums); j++ {
		t = append(t, nums[j])
		dfs(j+1, nums, t, res)
		t = t[:len(t)-1]
	}
}

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