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.

 public class Solution {
     public List<List<Integer>> permute(int[] nums) {
         Arrays.sort(nums);
         int length = nums.length;
         List<List<Integer>> result = new ArrayList<List<Integer>>();
         boolean[] visited = new boolean[length];

         dfs(nums, visited, new ArrayList<Integer>(), result);
         return result;
     }

     private void dfs(int[] nums, boolean[] visited, List<Integer> record, List<List<Integer>> result) {
         if (record.size() == nums.length) {
             if (!result.contains(record))
                 result.add(new ArrayList<Integer>(record));
             return;
         }
         for (int i = 0; i < nums.length; i++) {
             if (!visited[i]) {
                 record.add(nums[i]);
                 visited[i] = true;
                 dfs(nums, visited, record, result);
                 // Restore
                 record.remove(record.size() - 1);
                 visited[i] = false;
             }
         }
     }
 }

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

 class Solution:
     def dfs(self, nums: List[int], record: List[int], result: List[List[int]]) -> None:
         if not nums:
             result.append(record)
         for i in range(len(nums)):
             self.dfs(nums[:i] + nums[i + 1:], record + [nums[i]], result)

     def permute(self, nums: List[int]) -> List[List[int]]:
         result = []
         self.dfs(nums, [], result)
         return result