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