README_EN.md

78. Subsets

中文文档

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]]:
        def dfs(u, t):
            if u == len(nums):
                ans.append(t[:])
                return
            dfs(u + 1, t)
            t.append(nums[u])
            dfs(u + 1, t)
            t.pop()

        ans = []
        dfs(0, [])
        return ans

Java

class Solution {
    private List<List<Integer>> ans;
    private int[] nums;

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

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

C++

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

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

Go

func subsets(nums []int) [][]int {
	var ans [][]int
	var dfs func(u int, t []int)
	dfs = func(u int, t []int) {
		if u == len(nums) {
			ans = append(ans, append([]int(nil), t...))
			return
		}
		dfs(u+1, t)
		t = append(t, nums[u])
		dfs(u+1, t)
		t = t[:len(t)-1]
	}
	var t []int
	dfs(0, t)
	return ans
}

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