classSolution{publicbooleanisPowerOfTwo(intx){returnx >0&& (x ==1|| (x %2==0&& isPowerOfTwo(x /2))); } } Python 实现(递归) classSolution:defisPowerOfTwo(self, x):""" :type n: int :rtype: bool """returnx >0and(x ==1o
LeetCode power of 2 Given an integer, write a function to determine if it is a power of two. Example 1: Input: 1 Output: true Explanation: 20 = 1 Example 2: Input: 16 Output: true Explanation: 24 = 16 Example 3: Input: 218 Output: false classSolution {publicbooleanisPowerOfTwo(in...
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...
2)Power of 3. 是否为3的幂指数 Loading...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 ...
Leetcode上有种解法,就是让n&n-1,假设n仅仅含有一个1那么结果就是零。 简单解法: classSolution{public:boolisPowerOfTwo(intn){if(n<=0)returnfalse;intcount=0;while(n!=0){if(n%2!=0){count++;}n=n>>1;if(count>1)returnfalse;}returntrue;}}; ...
LeetCode 342: 4 的幂 时间复杂度:O(1) 只需要使用常数次位运算和布尔运算 空间复杂度:O(1) 只需要使用常数个额外变量即可 代码(Python3) class Solution: def isPowerOfTwo(self, n: int) -> bool: # 非正数一定不是 2 的幂次方 if n <= 0: return False return (n & (n - 1)) == 0 ...
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。
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?
[LeetCode] Power of Two, Power of Three, Power of Four 三道基本相同的题目,都可以用位操作、递归和迭代来做。 Power of Two 1. Bit Manipulation -- 2 ms beats 21.88% 2. Iteration -- 2 ms beats 21.88% Power of Three 1. Recursion -- 20 ms beats 24% 2. Iteration -- 15-17 ms ...
[i] 就是 words[j] 的一个子字符串。..." 的子字符串。...示例 2:输入:words = ["leetcode","et","code"] 输出:["et","code"] 解释:"et" 和 "code" 都是 "leetcode" 的子字符串。...解题先排序,按长度升序每个单词在后序的单词中查找 class Solution { public: vector stringMatching(...