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? public class Solution { public boolean i…

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. 个人博客:http://www.cnblogs.com/wdfwolf3/ 这道题本身没有难度,这里只是介绍两种思路,当我们判断出它二进制只有1个1的时候,即必为2的幂时,如何进一步判断它是…

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? 我的解法极其垃圾,建议不要看. public class Solution {…

Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example 1: Input: 16 Output: true Example 2: Input: 5 Output: false Follow up: Could you solve it without loops/recursion? 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…

public class Solution { public bool IsPowerOfFour(int num) { ) && ((num & (num - )) == ) && ((num & 0x55555555) == num); } } https://leetcode.com/problems/power-of-four/#/description…

This  is another  "Pick One" Problem :[Problem:342-Power of Four] Given an integer (signed bits), write a function to check whether it . Example: Given num = , , return false. Follow up: Could you solve it without loops/recursion? Python Codes:…

1.2的幂 正确写法: class Solution { public: bool isPowerOfTwo(int n) { ) return false; )) == ; } }; 错误写法1: &符号的短路原则,如果&前面为false了就不会计算后面的了 class Solution { public: bool isPowerOfTwo(int n) { ) return false; retur class Solution { public: bool isPowerOfFou…

位运算相关 三道题 231. Power of Two Given an integer, write a function to determine if it is a power of two. (Easy) 分析: 数字相关题有的可以考虑用位运算,例如&可以作为筛选器. 比如n & (n - 1) 可以将n的最低位1删除,所以判断n是否为2的幂,即判断(n & (n - 1) == 0) 代码: class Solution { public: bool isPowerOf…

Power of Four 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? /************************…