[Leetcode] 483. Smallest Good Base 解题报告

最新推荐文章于 2019-07-14 16:39:29 发布

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本文介绍了一种使用二分查找的方法来解决寻找字符串表示的大整数n的最小优秀基数的问题。优秀基数是指能使该整数在该基数下所有位均为1的基数。通过逐步减少位数并进行二分查找，可以有效地找到符合条件的最小基数。

题目：

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:

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

思路：

这道题目的基本思路是二分查找。注意到不同的k将会对应不同的长度d，使得11...1(一共d位)在k进制下表示n。而最小的k就意味着最大的d。所以我们就让d从大到小循环，看看11...1(一共d位)是否可以在某个base k下构成n。

对于长度为d的这个数，k的最小值显然是1，最大值则是power(tn, 1.0 / d) + 1。于是我们在这个区间内二分查找合适的k，并返回即可。

代码：

class Solution {
public:
    string smallestGoodBase(string n) {
        unsigned long long tn = static_cast<unsigned long long>(stoll(n));
        unsigned long long x = 1;
        for (int i = 62; i >= 1; --i) {
            if ((x << i) < tn) {
                unsigned long long cur = helper(tn, i);
                if (cur != 0) {
                    return to_string(cur);
                }
            }
        }
        return to_string(tn - 1);
    }
private:
    unsigned long long helper(unsigned long long n,int d) {
        double tn = static_cast<double>(n);
        unsigned long long right = static_cast<unsigned long long>(pow(tn, 1.0 / d) + 1);
        unsigned long long left = 1;
        while (left <= right){
            unsigned long long mid= left + (right - left) / 2;
            unsigned long long sum = 1,cur = 1;
            for (int i = 1;i <= d; ++i) {
                cur *= mid;
                sum += cur;
            }
            if (sum == n) {
                return mid;
            }
            if (sum > n) {
                right = mid - 1;
            }
            else {
                left = mid + 1;
            }
        }
        return 0;
    }
};