class Solution { public boolean isPowerOfTwo(int n) { return n>0 && Integer.bitCount(n) == 1; }} 因为只有一个 1，所以消除 1 后为 0.class Solution { public boolean isPowerOfTwo(int n) { return n>0 && (n&(n-1))==0; }} 8. 判断一个数是不是 4 的 n 次方...
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),写一个函数来检查它是不是4的次方数。
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. Returntrueif and only if we can do this in a way such that the resulting number is a power of 2. Example 1: Input:1Output:true Example ...
A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= pmax(u) of power, may consume an amount 0 <= c(u) <= min(s(u),cmax(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following ...
技巧5 判断是否为2的幂通过n与n-1进行与运算,结果为0则是2的幂 int num = 16;boolean isPowerOfTwo = (num & (num - 1)) == 0; 技巧6 位移操作左移(<<)和右移(>>)操作进行位移运算 int num = 8;int leftShifted = num << 1;int rightShifted = num >> 1; 技巧7 判断两个数是否异号...
bool isPowerOfTwo(int n) { return n > 0 && (n & -n) == n; //提取1,判断是还等于n,等于说明原来的数只有一个1 } }; 1. 2. 3. 4. 5. 6. 342. 4的幂【简单】 类似上一题 思路:n>0 && 1的位数只有一位 && 出现在奇数位上 ...
[LeetCode] 231. Power of Two 2的次方数 Given an integer, write a function to determine if it is a power of two. Example 1: Input: 1 Output: true Example 2: Input: 16 Output: true Example 3: Input: 218 Output: false 给一个整数,写一个函数来判断它是否为2的次方数。
Leetcode力扣 264 | 丑数 II Ugly Number II 21:00 Leetcode力扣 268 | 缺失数字 Missing Number 27:10 Leetcode力扣 287 | 寻找重复数 Find the Duplicate Number 10:42 字节大佬终于把python做成动画片了,通俗易懂,2023最新版,学完即可就业!拿走不谢,学不会我退出IT界!!
bool isPowerOfThree(int n){ if(n==0){ return false; } double x=log((double)n)/log(3.0); double e=1e-12; int ifloor=floor(x),iceil=ceil(x); if(x-ifloor<e||iceil-x<e){ return true; } return false; } 1. 2.
// #231 Description: Power of Two | LeetCode OJ 解法1:注意边界条件。 // Solution 1: Edge cases. 代码1 // Code 1 232 Implement Queue using Stacks // #232 用栈实现队列 描述:如题。 // #232 Description: Implement Queue using Stacks | LeetCode OJ 解法1:负负得正。 // Solution 1: ...