Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,[1,1,2] have the following unique permutations:[1,1,2], [1,2,1], and [2,1,1]
Tricky here is to avoid duplicated solutions.
1 public class Solution { 2 public List<List<Integer>> permuteUnique(int[] nums) { 3 Arrays.sort(nums); 4 int length = nums.length; 5 boolean[] visited = new boolean[length]; 6 List<List<Integer>> result = new ArrayList<List<Integer>>(); 7 8 dfs(nums, visited, new ArrayList<Integer>(), result); 9 return result; 10 } 11 12 private void dfs(int[] nums, boolean[] visited, List<Integer> record, List<List<Integer>> result) { 13 if (record.size() == nums.length) { 14 result.add(new ArrayList<Integer>(record)); 15 return; 16 } 17 int prev = -1; 18 for (int i = 0; i < nums.length; i++) { 19 if (visited[i]) 20 continue; 21 if (prev != -1 && nums[i] == nums[prev]) 22 continue; 23 record.add(nums[i]); 24 visited[i] = true; 25 dfs(nums, visited, record, result); 26 record.remove(record.size() - 1); 27 visited[i] = false; 28 prev = i; 29 } 30 } 31 }
Python3
class Solution: def dfs(self, nums: List[int], visited: List[int], record: List[int], result: List[List[int]]) -> None: if (len(record) == len(nums)): if record not in result: result.append(copy.deepcopy(record)) return for i in range(len(nums)): if visited[i]: continue # if previous node equals to current node, and previous one has not been visited, ignore dfs for current node. if i > 0 and nums[i] == nums[i - 1] and not visited[i - 1]: continue visited[i] = True record.append(nums[i]) self.dfs(nums, visited, record, result) # revert status visited[i] = False record.pop() def permuteUnique(self, nums: List[int]) -> List[List[int]]: nums.sort() result = [] visited = [False] * len(nums) self.dfs(nums, visited, [], result) return result
原文：https://www.cnblogs.com/ireneyanglan/p/4884946.html