Permutations 解答

Question

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

Solution

Traditional backtracking way to solve this problem.

 1 public class Solution {
 2     public List<List<Integer>> permute(int[] nums) {
 3         Arrays.sort(nums);
 4         int length = nums.length;
 5         List<List<Integer>> result = new ArrayList<List<Integer>>();
 6         boolean[] visited = new boolean[length];
 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             if (!result.contains(record))
15                 result.add(new ArrayList<Integer>(record));
16             return;
17         }
18         for (int i = 0; i < nums.length; i++) {
19             if (!visited[i]) {
20                 record.add(nums[i]);
21                 visited[i] = true;
22                 dfs(nums, visited, record, result);
23                 // Restore
24                 record.remove(record.size() - 1);
25                 visited[i] = false;
26             }
27         }
28     }
29 }

Another solution is that we don't need to store visited status, but we just need to modify "nums" object.

 1 class Solution:
 2     def dfs(self, nums: List[int], record: List[int], result: List[List[int]]) -> None:
 3         if not nums:
 4             result.append(record)
 5         for i in range(len(nums)):
 6             self.dfs(nums[:i] + nums[i + 1:], record + [nums[i]], result)
 7     
 8     def permute(self, nums: List[int]) -> List[List[int]]:
 9         result = []
10         self.dfs(nums, [], result)
11         return result