---

comments: true
edit_url: https://github.com/doocs/leetcode/edit/main/lcof2/%E5%89%91%E6%8C%87%20Offer%20II%20084.%20%E5%90%AB%E6%9C%89%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0%E9%9B%86%E5%90%88%E7%9A%84%E5%85%A8%E6%8E%92%E5%88%97/README.md

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]*ndef dfs(i):if len(path)==n:res.append(path[:])returnif len(path)>n:returnfor 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]=1dfs(j+1)path.pop()used[j]=0dfs(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] = truedfs(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] = truedfs(u + 1, n, nums, &used, &path, &res)used[i] = falsepath.removeLast()}}
}