753. Cracking the Safe

There is a box protected by a password. The password is n digits, where each letter can be one of the first k digits 0, 1, ..., k-1.

You can keep inputting the password, the password will automatically be matched against the last n digits entered.

For example, assuming the password is "345", I can open it when I type "012345", but I enter a total of 6 digits.

Please return any string of minimum length that is guaranteed to open the box after the entire string is inputted.

Example 1:

Input: n = 1, k = 2
Output: "01"
Note: "10" will be accepted too.

Example 2:

Input: n = 2, k = 2
Output: "00110"
Note: "01100", "10011", "11001" will be accepted too.

Note:

1. n will be in the range [1, 4].
2. k will be in the range [1, 10].
3. k^n will be at most 4096.

class Solution:
    def crackSafe(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: str
        """
        seen = set()
        ans = []
        def dfs(node):
            for x in map(str, range(k)):
                nei = node + x
                if nei not in seen:
                    seen.add(nei)
                    dfs(nei[1:])
                    ans.append(x)
        dfs("0" * (n-1))
        return "".join(ans) + "0" * (n-1)

思路是利用欧拉回路的方法，使用dfs走遍所有节点。