326. 3的幂 给定一个整数,写一个函数来判断它是否是 3 的幂次方. 示例 1: 输入: 27 输出: true 示例 2: 输入: 0 输出: false 示例 3: 输入: 9 输出: true 示例 4: 输入: 45 输出: false 进阶: 你能不使用循环或者递归来完成本题吗? class Solution { public boolean isPowerOfThree(int n) { if (n == 0) { return false; } while (n % 3 ==…

给定一个整数,写一个函数来判断它是否是 3 的幂次方. 示例 1: 输入: 27 输出: true 示例 2: 输入: 0 输出: false 示例 3: 输入: 9 输出: true 示例 4: 输入: 45 输出: false 进阶: 你能不使用循环或者递归来完成本题吗? 思路 循环/3看最后是不是等于1就好了 代码 class Solution(object): def isPowerOfThree(self, n): """ :type n: int :rtype:…

342. 4的幂 给定一个整数 (32 位有符号整数),请编写一个函数来判断它是否是 4 的幂次方. 示例 1: 输入: 16 输出: true 示例 2: 输入: 5 输出: false 进阶: 你能不使用循环或者递归来完成本题吗? class Solution { public boolean isPowerOfFour(int num) { int x = 0x55555555; return (num > 0)&&((num&(num-1))==0)&((nu…

231. 2的幂 给定一个整数,编写一个函数来判断它是否是 2 的幂次方. 示例 1: 输入: 1 输出: true 解释: 20 = 1 示例 2: 输入: 16 输出: true 解释: 24 = 16 示例 3: 输入: 218 输出: false PS: 2的次幂和他的上一位数&的结果为0 8的二进制就是1000 7的二进制就是0111 结果========0000 class Solution { public boolean isPowerOfTwo(int n) { if(n <…

leetcode 326. Power of Three(不用循环或递归) Given an integer, write a function to determine if it is a power of three. Follow up: Could you do it without using any loop / recursion? 题意是判断一个数是否是3的幂,最简单的也最容易想到的办法就是递归判断,或者循环除. 有另一种方法就是,求log以3为底n的对数.类似 如果n=9,则…

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