【leetcode】1022. Smallest Integer Divisible by K

题目如下：

Given a positive integer K, you need find the smallest positive integer N such that N is divisible by K, and N only contains the digit 1.

Return the length of N.  If there is no such N, return -1.

Example 1:

Input: 1
Output: 1
Explanation: The smallest answer is N = 1, which has length 1.

Example 2:

Input: 2
Output: -1
Explanation: There is no such positive integer N divisible by 2.

Example 3:

Input: 3
Output: 3
Explanation: The smallest answer is N = 111, which has length 3.

Note:

• 1 <= K <= 10^5

解题思路：题目要求找出最小的一个X,使得K*X = 111.....111，一开始我的思路是找出这样的X，例如K=19937，那么X的最后一位一定是3，接下来再计算X的倒数第二位，但是这样会超时，可能是掉入了死循环。其实反过来想想，我们可以111.....111去除以K，判断能否被K整除。首先找出大于或等于K的最小的11...11，然后除以K得到余数，余数后面继续加上最少数量的1使得余数大于K，然后再除以K直到余数为0为止。对于无法被整除的情况，经过若干次操作之后，一定会出现重复的余数，只要用字典记录出现过的余数，如果有重复出现则返回-1。

代码如下：

class Solution(object):
    def smallestRepunitDivByK(self, K):
        """
        :type K: int
        :rtype: int
        """
        len_k = len(str(K))
        res = len_k
        div = '' * (len_k)
        if int(div) < K:
            div += ''
            res += 1
        dic_remainder = {}
        while True:
            remainder = int(div) % K
            if remainder == 0:
                return res
            elif remainder in dic_remainder:
                return -1
            dic_remainder[remainder] = 1
            sr = str(remainder)
            div = sr + '' * (len_k - len(sr))
            res += (len_k - len(sr))
            if int(div) < K:
                div += ''
                res += 1