0326. Power of Three

# 326. Power of Three#

## 题目 #

Given an integer, write a function to determine if it is a power of three.

Example 1:

``````Input: 27
Output: true
``````

Example 2:

``````Input: 0
Output: false
``````

Example 3:

``````Input: 9
Output: true
``````

Example 4:

``````Input: 45
Output: false
``````

Could you do it without using any loop / recursion?

## 解题思路 #

• 判断一个数是不是 3 的 n 次方。
• 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。

## 代码 #

``````
package leetcode

// 解法一 数论
func isPowerOfThree(n int) bool {
// 1162261467 is 3^19,  3^20 is bigger than int
return n > 0 && (1162261467%n == 0)
}

// 解法二 打表法
func isPowerOfThree1(n int) bool {
// 1162261467 is 3^19,  3^20 is bigger than int
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}
_, ok := allPowerOfThreeMap[n]
return ok
}

// 解法三 循环
func isPowerOfThree2(num int) bool {
for num >= 3 {
if num%3 == 0 {
num = num / 3
} else {
return false
}
}
return num == 1
}

``````

Apr 8, 2023