Leetcode 483. Smallest Good Base

For an integer n, we call k>=2 a good base of n, if all digits of n base k are 1.

Now given a string representing n, you should return the smallest good base of n in string format. 

Example 1:

Input: "13"
Output: "3"
Explanation: 13 base 3 is 111.

Example 2:

Input: "4681"
Output: "8"
Explanation: 4681 base 8 is 11111.

Example 3:

Input: "1000000000000000000"
Output: "999999999999999999"
Explanation: 1000000000000000000 base 999999999999999999 is 11.

Note:

1. The range of n is [3, 10^18].
2. The string representing n is always valid and will not have leading zeros.

Next challenges: Palindrome Number Best Meeting Point Encode and Decode TinyURL 

思路：首先完成字符串到数字的转换，然后对于特定的num，当前的最长的其他进制的表示长度是进制为2时的表示长度，就是log₂(num)+1（注意：10..0有t个0，那么10..0=2^t）。那么指数i的遍历区间就是[1，log₂(num)+1]。对于当前数的当前指数i，可能的base整数取值是num^(1/(i-1))。

代码：

public class Solution {
    public String smallestGoodBase(String n) {
        long num = Long.parseLong(n);
        int maxIndex = (int) (Math.log(num)/Math.log(2) + 1);
        for(int i = maxIndex; i >= 3; i--) {
            long base = (long)Math.pow(num, 1.0 / (i - 1)), sum = 1, cur = 1;
            for(int j = 1; j < i; j++) {
                sum += (cur *= base);
            }
            if(sum == num) return String.valueOf(base);
            // java中没有无符号数，而且Math.pow(base, i) - 1存在精度问题，例如样例"14919921443713777"，结果应该为"496"，但是下列语句的结果是"14919921443713776"
            // if((long)((Math.pow(base, i) - 1) / (base - 1)) == num) return String.valueOf(base);
        }
        return String.valueOf(num - 1);
    }
}