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? Credits:Special thanks to @yukuairoy fo…

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? Credits:Special thanks to @dietpepsi for adding this problem and creating all test cases. 这道题让我们判断一个数是不是3的次方数,在Le…

Given an integer, write a function to determine if it is a power of two. Hint: Could you solve it in O(1) time and using O(1) space? 这道题让我们判断一个数是否为2的次方数,而且要求时间和空间复杂度都为常数,那么对于这种玩数字的题,我们应该首先考虑位操作 Bit Operation.在LeetCode中,位操作的题有很多,比如比如Repeated DNA Seque…

原题链接在这里:https://leetcode.com/problems/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…

原题链接在这里:https://leetcode.com/problems/power-of-three/ 与Power of Two类似.检查能否被3整除,然后整除,再重复检查结果. Time Complexity: O(logn). n是原始数字.Space: O(1). AC Java: public class Solution { public boolean isPowerOfThree(int n) { if(n < 1){ return false; } while(n > 1…

Given an integer, write a function to determine if it is a power of two. 题目意思: 给定一个整数,判断是否是2的幂 解题思路: 如果一个整数是2的幂,则二进制最高位为1,减去1后则最高位变为0,后面全部是1,相与判读是否为零,注意负数和0,负数的最高位是1. 更多二进制的问题可以参考<编程之美>中二进制有多少个1,面试时容易问. 源代码: class Solution { public: bool isPowerOfTw…

Given an integer, write a function to determine if it is a power of two. Hint: Could you solve it in O(1) time and using O(1) space? 最容易想到的方法是用n反复的去除以2,一直到n变成1.这期间如果出现不能被2整除的数,那么一开始的n就不是2的power. class Solution(object): def isPowerOfTwo(self, n): "&qu…

class Solution { public: double pow(double x, int n) { double a=1; if(n==0)return 1; if(x==1)return 1; if(x==-1) { if(n&1)return -1; else return 1; } int absn=abs(n); a=power(x,absn); if(n<0) { a=1/a; } return a; } double power(double x,int n) { do…

Description: Given an integer, write a function to determine if it is a power of two. public class Solution { public boolean isPowerOfTwo(int n) { boolean flag = false; if(n == 0) return false; while(true) { if(n == 1) { flag = true; break; } if(n %…

题目:Power of Two Given an integer, write a function to determine if it is a power of two. 题意:判断一个数是否是2的k次方数. 思路: 2的次方数使用2进制表示则必然只有最高位是1其他都是0: 这样判断一个数最多需要循环32次就能得出结果. 则程序如下: bool isPowerOfTwo(int n){ if (!n)return false; while ((n&0x01) != 0x01){//2的次方…