n&(n-1);//n-1会将n中从小到大第一个为1的位置变成0,这样就能判断只有一个1的情况。 boolisPowerOfTwo(intn){//2的幂只有一个1,则使用n&(n-1)来统计1的个数returnn >0&& !(n & (n -1)); } 题目:Power of Three Given an integer, write a function to determine if it is a power ...
整理自leetCode 231.Power of Two && 236.Power of Three Given an integer, write a function to determine if it is a power of two. 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? 常规解题手段:...
LeetCode*326. Power of Three LeetCode题目链接 注意:凡是以英文出现的,都是题目提供的,包括答案代码里的前几行。 题目: 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? 解析: 这一题最开始的想......
Can you solve this real interview question? Power of Two - Given an integer n, return true if it is a power of two. Otherwise, return false. An integer n is a power of two, if there exists an integer x such that n == 2x. Example 1: Input: n = 1 Ou
Leetcode力扣239.滑动窗口最大值 丶mars_ 3481 3 Leetcode 231. Power of Two (Python) Myron_yoo 21 0 Leetcode刷题 2 两数相加 Add Two Numbers JS老毕 2257 4 Leetcode力扣22 手画图解版|括号生成Generate Parentheses 爱学习的饲养员 4029 4 一周刷爆LeetCode,算法大神左神(左程云)耗时100天打造...
2)Power of 3. 是否为3的幂指数 Loading...leetcode.com/problems/power-of-three/ 2.1)采用循环 class Solution: def isPowerOfThree(self, n: int) -> bool: if not n: return False while n: if n==1: return True if n%3: return False n //=3 return True ...
LeetCode(65)-Power of Four 题目:Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example: Given num = 16, return true. Given num = 5, return false. Follow up: Could you solve it without loops/recursion?
LeetCode算法题-Power of Four(Java实现-六种解法) 这是悦乐书的第205次更新,第216篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第72题(顺位题号是342)。给定一个整数(带符号的32位),写一个函数来检查它是否为4的幂。例如: 输入:16...
LeetCode笔记:342. Power of Four 问题: Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example: Given num = 16, return true. Given num = 5, return false. Follow up: Could you solve it without loops/recursion?
这道题让我们判断一个数是不是3的次方数,在LeetCode中,有一道类似的题目Power of Two,那道题有个非常简单的方法,由于2的次方数实在太有特点,最高位为1,其他位均为0,所以特别容易,而3的次方数没有显著的特点,最直接的方法就是不停地除以3,看最后的余数是否为1,要注意考虑输入是负数和0的情况,参见代码如下...