Question

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

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

    def dfs(u):
      if u == n:
        res.append(path.copy())
        return
      for i in range(n):
        if used[i] or (i > 0 and nums[i] == nums[i - 1] and not used[i - 1]):
          continue
        path[u] = nums[i]
        used[i] = True
        dfs(u + 1)
        used[i] = False

    dfs(0)
    return res

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);
    }
  }
}

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;
    }
  }
};

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
  }
}

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;
      }
    }
  }
}