学习了一个更简洁的(https://leetcode.com/discuss/45017/5-lines-o-1-space%26time-c-solution-no-hash) 自己优化了一下 一行代码AC。 classSolution {public:boolisPowerOfTwo(intn) {returnn<=0?false: !(n & (n-1)); } };
classSolution(object): defisPowerOfTwo(self, n): """ :type n: int :rtype: bool """ #-4??? c=0 whilen >0: ifn &1: c+=1 n >>=1 returnc==1 标签:leetcode 好文要顶关注我收藏该文微信分享 bonelee 粉丝-904关注 -1
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:...
题目地址:https://leetcode.com/problems/reordered-power-of-2/description/题目描述:Starting with a positive integer N, we reorder the digits in any order (including the original order) such that the leading digit is not zero.Return true if and only if we can do this in a way such that ...
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 Runtime: 80 ms, faster than 52.08% of Python3 online submissions...
AI代码解释 publicclassSolution{publicbooleanisPowerOfFour(int num){// 转化为二进制数来判断if(num<0)returnfalse;String binaryStr=Integer.toBinaryString(num);for(int i=0;i<binaryStr.length();i++){if(i==0&&binaryStr.charAt(i)!='1')returnfalse;elseif(i>0&&binaryStr.charAt(i)!='0')...
classSolution {public:boolisPowerOfThree(intn) {return(n >0&∫(log10(n) / log10(3)) - log10(n) / log10(3) ==0); }}; 类似题目: Power of Two 参考资料: https://leetcode.com/discuss/78532/summary-all-solutions-new-method-included-at-15-30pm-jan-8th...
powerof two power-of-two class Solution { public: bool isPowerOfTwo(int n) { return n>=1 && !(n&(n-1)); } }; n=10000***000, <=> n&(n-1)=0 是这种方法的核心 https://leetcod i++ 版本号 c++ 链表 转载 mob604756f261ee ...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算法解决。大家加油!:) 0 stars 721 forks ...
解题:是2的次幂的整数为:1、2、4、8、16...其二进制表示都是1000...,n-1的二进制表示为0111...将n于n-1使用按位与(&)操作,看结果是否为0即可判断。 复杂度:时间复杂度:O(1),空间复杂度:O(1)。 classSolution{public:boolisPowerOfTwo(intn){if(n<=0)returnfalse;return!(n&(n-1));}};...