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77 lines
1.6 KiB
Markdown
Executable File
77 lines
1.6 KiB
Markdown
Executable File
# [326. Power of Three](https://leetcode.com/problems/power-of-three/)
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## 题目
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Given an integer, write a function to determine if it is a power of three.
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**Example 1**:
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Input: 27
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Output: true
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**Example 2**:
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Input: 0
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Output: false
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**Example 3**:
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Input: 9
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Output: true
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**Example 4**:
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Input: 45
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Output: false
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**Follow up**:
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Could you do it without using any loop / recursion?
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## 题目大意
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给定一个整数，写一个函数来判断它是否是 3 的幂次方。
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## 解题思路
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- 判断一个数是不是 3 的 n 次方。
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- 这一题最简单的思路是循环，可以通过。但是题目要求不循环就要判断，这就需要用到数论的知识了。由于 3^20 超过了 int 的范围了，所以 3^19 次方就是 int 类型中最大的值。这一题和第 231 题是一样的思路。
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## 代码
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```go
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package leetcode
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// 解法一 数论
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func isPowerOfThree(n int) bool {
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// 1162261467 is 3^19, 3^20 is bigger than int
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return n > 0 && (1162261467%n == 0)
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}
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// 解法二 打表法
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func isPowerOfThree1(n int) bool {
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// 1162261467 is 3^19, 3^20 is bigger than int
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allPowerOfThreeMap := map[int]int{1: 1, 3: 3, 9: 9, 27: 27, 81: 81, 243: 243, 729: 729, 2187: 2187, 6561: 6561, 19683: 19683, 59049: 59049, 177147: 177147, 531441: 531441, 1594323: 1594323, 4782969: 4782969, 14348907: 14348907, 43046721: 43046721, 129140163: 129140163, 387420489: 387420489, 1162261467: 1162261467}
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_, ok := allPowerOfThreeMap[n]
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return ok
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}
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// 解法三 循环
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func isPowerOfThree2(num int) bool {
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for num >= 3 {
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if num%3 == 0 {
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num = num / 3
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} else {
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return false
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}
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}
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return num == 1
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}
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``` |