LeetCode笔记:342. 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? 大意: 给出一个数字(有符号32位),写...
publicclassSolution{publicbooleanisPowerOfFour(intnum){returnnum >0&& (num & (num -1)) ==0&& (num &0x55555555) !=0; } } 函数法 判断取以4为底的log之后,强转成int,再取上4的幂是不是原来的数字。 classSolution(object):defisPowerOfFour(self, num):""" :type num: int :rtype: bool ...
classSolution {public:boolisPowerOfFour(intnum) {returnnum >0&& !(num & (num -1)) && (num -1) %3==0; } }; 类似题目: Power of Three Power of Two 参考资料: https://leetcode.com/problems/power-of-four/ https://leetcode.com/problems/power-of-four/discuss/80457/Java-1-line-(c...
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 342: Power of Four 题目描述 给定一个整数,编写一个函数以确定它是否为4的幂。 解题思路 4的整数次幂为整数并且它的的二进制表示有且只有一个 奇数位为1。 AC代码...LeetCode #342 - Power of Four 题目描述: Given an integer (signed 32 bits), write a function to check whether it is...
(n & 0x55555555) != 0: 判断 n 的二进制位是否存在偶数位为 1 // // 当前仅当上述两个条件都满足时, n 是 4 的幂次方 return (n & (n - 1)) == 0 && n & 0x55555555 != 0 } 题目链接: Power of Four: leetcode.com/problems/p 计算布尔二叉树的值: leetcode.cn/problems/po Leet...
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...
Power of Two Given an integer, write a function to determine if it is a power of two. 整除法 复杂度 时间O(1) 空间 O(1) 思路 最简单的解法,不断将原数除以2,一旦无法整除,余数不为0,则说明不是2的幂,如果整除到1,说明是2的幂。
详见:https://leetcode.com/problems/power-of-four/description/ C++: 方法一: classSolution{public:boolisPowerOfFour(intnum){while(num&&(num%4==0)){num/=4;}returnnum==1;}}; 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 方法二:
3 return (Math.log10(n) / Math.log10(3)) % 1 == 0; 4 } 5 } 1. 2. 3. 4. 5. 1 class Solution { 2 public boolean isPowerOfFour(int num) { 3 return Math.log(num) / Math.log(4) % 1 == 0; 4 } 5 } 1. 2. 3. 4. 5. LeetCode 题目总结...