README_EN.md

90. Subsets II

中文文档

Description

Given an integer array nums that may contain duplicates, 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 = [1,2,2]
Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]

Example 2:

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

 

Constraints:

• 1 <= nums.length <= 10
• -10 <= nums[i] <= 10

Solutions

Python3

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        def dfs(u, t):
            ans.append(t[:])
            for i in range(u, len(nums)):
                if i != u and nums[i] == nums[i - 1]:
                    continue
                t.append(nums[i])
                dfs(i + 1, t)
                t.pop()

        ans = []
        nums.sort()
        dfs(0, [])
        return ans

Java

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

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

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

C++

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> ans;
        vector<int> t;
        dfs(0, t, nums, ans);
        return ans;
    }

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

Go

func subsetsWithDup(nums []int) [][]int {
	sort.Ints(nums)
	var ans [][]int
	var dfs func(u int, t []int)
	dfs = func(u int, t []int) {
		ans = append(ans, append([]int(nil), t...))
		for i := u; i < len(nums); i++ {
			if i != u && nums[i] == nums[i-1] {
				continue
			}
			t = append(t, nums[i])
			dfs(i+1, t)
			t = t[:len(t)-1]
		}
	}
	var t []int
	dfs(0, t)
	return ans
}

TypeScript

function subsetsWithDup(nums: number[]): number[][] {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const t: number[] = [];
    const res: number[][] = [];
    const dfs = (i: number) => {
        if (i === n) {
            res.push([...t]);
            return;
        }
        t.push(nums[i]);
        dfs(i + 1);
        const num = t.pop();
        while (i < n && nums[i] == num) {
            i++;
        }
        dfs(i);
    };
    dfs(0);
    return res;
}

Rust

impl Solution {
    fn dfs(mut i: usize, t: &mut Vec<i32>, res: &mut Vec<Vec<i32>>, nums: &Vec<i32>) {
        let n = nums.len();
        if i == n {
            res.push(t.clone());
            return;
        }
        t.push(nums[i]);
        Self::dfs(i + 1, t, res, nums);
        let num = t.pop().unwrap();
        while i < n && num == nums[i] {
            i += 1;
        }
        Self::dfs(i, t, res, nums);
    }

    pub fn subsets_with_dup(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
        nums.sort();
        let mut res = Vec::new();
        Self::dfs(0, &mut Vec::new(), &mut res, &nums);
        res
    }
}

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