题目: Given an integer, write a function to determine if it is a power of two. 分析: 利用cpp函数log即可,注意当integer==0时的特殊情况。 代码: class Solution {public:bool isPowerOfTwo(int n) {if (n==0)return fa
Given an integer, write a function to determine if it is a power of two. 解题思路 牛逼的解法,不解释,直接看代码 C++代码 class Solution {public:bool isPowerOfTwo(int n) {if(n <= 0) return false;return !(n & (n-1));
Given an integer, write a function to determine if it is a power of two. 题目链接:https://leetcode.com/problems/power-of-two/ 题目分析:大于0且二进制中只有一位为1 public class Solution {public boolean isPowerOfTwo
Problem: 231. 2 的幂 文章目录 思路解题方法Code 思路 说白了就是靠硬算,但是要知道对sum不进行控制就会导致直接超标,所以要在for循环的条件中加上sum <= n 解题方法 由思路可知 Code bool isPowerOfTwo(int n) {long int sum = 1;for(int i = 0; i <= n &&
题目 Given an integer, write a function to determine if it is a power of two. 代码 /*---------------------------------------* 日期:2015-08-02* 作者:SJF0115* 题目: 231.Power of Two* 网址:https://lee
Given an integer, write a function to determine if it is a power of two. 麻烦一点的话,用循环来判断。 class Solution {public:bool isPowerOfTwo(int n) {if(n<=0)return false;while(n){if(n%2!=0){if(n==1)return true