剑指 Offer II 084. 含有重复元素集合的全排列

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剑指 Offer II 084. 含有重复元素集合的全排列

题目描述

给定一个可包含重复数字的整数集合 nums ，按任意顺序 返回它所有不重复的全排列。

示例 1：

输入：nums = [1,1,2]
输出：
[[1,1,2],
 [1,2,1],
 [2,1,1]]

示例 2：

输入：nums = [1,2,3]
输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

提示：

• 1 <= nums.length <= 8
• -10 <= nums[i] <= 10

注意：本题与主站 47 题相同： https://leetcode.cn/problems/permutations-ii/

解法

方法一

Python3

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        n=len(nums)
        res=[]
        path=[]
        used=[0]*n

        def dfs(i):
            if len(path)==n:
                res.append(path[:])
                return
            if len(path)>n:
                return

            for j in range(n):
                if used[j]:continue #纵向避免重复选择（这题只能通过位置标记）

                #j and not used[j-1]很巧妙（既实现了同层剪枝，也避免了对异层【递归方向】的影响【可以选同值异位元素】）：起始位置之后的重复元素需要被跳过，而不同层（比如下一层递归）中的同值异位元素是可以使用的 
                # (j and not used[j-1]) 类似  重复元素组合问题中的j>i,只不过排列可以选到之前的元素
                if (j and not used[j-1]) and nums[j]==nums[j-1]:continue 
                
                path.append(nums[j])
                used[j]=1
                dfs(j+1)
                path.pop()
                used[j]=0
        dfs(0)
        return res

Java

class Solution {
    public List<List<Integer>> permuteUnique(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        int n = nums.length;
        boolean[] used = new boolean[n];
        Arrays.sort(nums);
        dfs(0, n, nums, used, path, res);
        return res;
    }

    private void dfs(
        int u, int n, int[] nums, boolean[] used, List<Integer> path, List<List<Integer>> res) {
        if (u == n) {
            res.add(new ArrayList<>(path));
            return;
        }
        for (int i = 0; i < n; ++i) {
            if (used[i] || (i > 0 && nums[i] == nums[i - 1] && !used[i - 1])) {
                continue;
            }
            path.add(nums[i]);
            used[i] = true;
            dfs(u + 1, n, nums, used, path, res);
            used[i] = false;
            path.remove(path.size() - 1);
        }
    }
}

C++

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> res;
        vector<int> path(n, 0);
        vector<bool> used(n, false);
        sort(nums.begin(), nums.end());
        dfs(0, n, nums, used, path, res);
        return res;
    }

    void dfs(int u, int n, vector<int>& nums, vector<bool>& used, vector<int>& path, vector<vector<int>>& res) {
        if (u == n) {
            res.emplace_back(path);
            return;
        }
        for (int i = 0; i < n; ++i) {
            if (used[i] || (i > 0 && nums[i] == nums[i - 1] && !used[i - 1])) continue;
            path[u] = nums[i];
            used[i] = true;
            dfs(u + 1, n, nums, used, path, res);
            used[i] = false;
        }
    }
};

Go

func permuteUnique(nums []int) [][]int {
	n := len(nums)
	res := make([][]int, 0)
	path := make([]int, n)
	used := make([]bool, n)
	sort.Ints(nums)
	dfs(0, n, nums, used, path, &res)
	return res
}

func dfs(u, n int, nums []int, used []bool, path []int, res *[][]int) {
	if u == n {
		t := make([]int, n)
		copy(t, path)
		*res = append(*res, t)
		return
	}
	for i := 0; i < n; i++ {
		if used[i] || (i > 0 && nums[i] == nums[i-1] && !used[i-1]) {
			continue
		}
		path[u] = nums[i]
		used[i] = true
		dfs(u+1, n, nums, used, path, res)
		used[i] = false
	}
}

C#

using System.Collections.Generic;
using System.Linq;

public class Solution {
    public IList<IList<int>> PermuteUnique(int[] nums) {
        var results = new List<IList<int>>();
        var temp = new List<int>();
        var count = nums.GroupBy(n => n).ToDictionary(g => g.Key, g => g.Count());
        Search(count, temp, results);
        return results;
    }

    private void Search(Dictionary<int, int> count, IList<int> temp, IList<IList<int>> results)
    {
        if (!count.Any() && temp.Any())
        {
            results.Add(new List<int>(temp));
            return;
        }
        var keys = count.Keys.ToList();
        foreach (var key in keys)
        {
            temp.Add(key);
            --count[key];
            if (count[key] == 0) count.Remove(key);
            Search(count, temp, results);
            temp.RemoveAt(temp.Count - 1);
            if (count.ContainsKey(key))
            {
                ++count[key];
            }
            else
            {
                count[key] = 1;
            }
        }
    }
}

Swift

class Solution {
    func permuteUnique(_ nums: [Int]) -> [[Int]] {
        var res = [[Int]]()
        var path = [Int]()
        var used = [Bool](repeating: false, count: nums.count)
        let sortedNums = nums.sorted()
        dfs(0, sortedNums.count, sortedNums, &used, &path, &res)
        return res
    }

    private func dfs(
        _ u: Int, _ n: Int, _ nums: [Int], _ used: inout [Bool], _ path: inout [Int], _ res: inout [[Int]]
    ) {
        if u == n {
            res.append(path)
            return
        }
        for i in 0..<n {
            if used[i] || (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {
                continue
            }
            path.append(nums[i])
            used[i] = true
            dfs(u + 1, n, nums, &used, &path, &res)
            used[i] = false
            path.removeLast()
        }
    }
}