What is the main pattern in Hexadecimal and Hexatrigesimal Conversion?

The core pattern is Math plus String: compute n squared and n cubed, then convert each value with repeated division and remainder, using the correct digit table for base 16 and base 36.

Why is a generic toBase function better here?

This problem asks for two numeral systems, but the conversion steps are identical except for the base and allowed symbols. One helper reduces branching and makes it less likely you will map digits inconsistently between hexadecimal and hexatrigesimal.

How do I handle digits above 9 in base 36?

Map values 10 through 35 to uppercase letters A through Z. A digit table string such as 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ lets you index directly by remainder.

What explains the output A91P1 when n = 13?

First, 13 squared is 169, and 169 in hexadecimal is A9. Then 13 cubed is 2197, and 2197 in base 36 is 1P1. Concatenating them gives A91P1.

Do I need to optimize beyond basic conversion loops for LeetCode 3602?

No. The constraint is small, so the important part is correct conversion and concatenation. Most wrong answers come from digit order, uppercase handling, or converting the wrong power, not from time complexity.

#3602

Easy

auto_awesome数学·string

LeetCode 题解工作台

十六进制和三十六进制转化

给你一个整数 n 。返回 n 2 的十六进制表示和 n 3 的三十六进制表示拼接成的字符串。十六进制数定义为使用数字 0 – 9 和大写字母 A - F 表示 0 到 15 的值。三十六进制数定义为使用数字 0 – 9 和大写字母 A - Z 表示 0 到 35 的值。示例 1：…

数学字符串

题目描述

给你一个整数 n。

返回 n² 的 十六进制表示 和 n³ 的 三十六进制表示 拼接成的字符串。

十六进制 数定义为使用数字 0 – 9 和大写字母 A - F 表示 0 到 15 的值。

三十六进制 数定义为使用数字 0 – 9 和大写字母 A - Z 表示 0 到 35 的值。

示例 1：

输入：n = 13

输出： "A91P1"

解释：

n² = 13 * 13 = 169。在十六进制中，它转换为 (10 * 16) + 9 = 169，对应于 "A9"。
n³ = 13 * 13 * 13 = 2197。在三十六进制中，它转换为 (1 * 36²) + (25 * 36) + 1 = 2197，对应于 "1P1"。
连接两个结果得到 "A9" + "1P1" = "A91P1"。

示例 2：

输入：n = 36

输出："5101000"

解释：

n² = 36 * 36 = 1296。在十六进制中，它转换为 (5 * 16²) + (1 * 16) + 0 = 1296，对应于 "510"。
n³ = 36 * 36 * 36 = 46656。在三十六进制中，它转换为 (1 * 36³) + (0 * 36²) + (0 * 36) + 0 = 46656，对应于 "1000"。
连接两个结果得到 "510" + "1000" = "5101000"。

提示:

1 <= n <= 1000

lightbulb

解题思路

方法一：模拟

我们定义一个函数 $\textit{f}(x, k)$ ，它将整数 $x$ 转换为以 $k$ 进制表示的字符串。该函数通过不断取模和整除来构建结果字符串。

对于给定的整数 $n$ ，我们计算 $n^2$ 和 $n^3$ ，然后分别将它们转换为十六进制和三十六进制字符串。最后，将这两个字符串连接起来返回。

时间复杂度 $O(\log n)$ ，空间复杂度 $O(1)$ 。

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class Solution:
    def concatHex36(self, n: int) -> str:
        def f(x: int, k: int) -> str:
            res = []
            while x:
                v = x % k
                if v <= 9:
                    res.append(str(v))
                else:
                    res.append(chr(ord("A") + v - 10))
                x //= k
            return "".join(res[::-1])

        x, y = n**2, n**3
        return f(x, 16) + f(y, 36)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
You notice the problem is really two base conversions glued together, not a tricky math identity.
question_mark
You suggest a generic toBase(x, b) helper with a 0-9 and A-Z digit table instead of hardcoding hexadecimal and base 36 separately.
question_mark
You explicitly guard against the order bug: convert n squared to hexadecimal first, then n cubed to hexatrigesimal.

warning

常见陷阱

外企场景

error
Using lowercase letters or the wrong digit table, which breaks the required uppercase output for both bases.
error
Forgetting to reverse the collected digits after repeated modulus and division, producing the representation backward.
error
Mixing up the powers or output order, such as converting n cubed to base 16 or concatenating the base 36 part first.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return the two converted strings as separate fields instead of one concatenated result, which removes the final join but keeps the same helper.
arrow_right_alt
Change the bases to another pair such as binary and base 7, which tests whether the conversion logic is truly generic.
arrow_right_alt
Allow n = 0, which adds the classic edge case where toBase must return 0 instead of an empty string.

help

常见问题

外企场景

继续练习

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