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 ...
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 手把手带你刷Leetcode力扣|各个击破数据结构和算法|大厂面试必备技能【已完结】 ...
check-if-binary-string-has-at-most-one-segment-of-ones check-if-it-is-a-good-array check-if-n-and-its-double-exist check-if-number-has-equal-digit-count-and-digit-value check-if-number-is-a-sum-of-powers-of-three check-if-numbers-are-ascending-in-a-sentence check-if-object...
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 ...
LeetCode笔记:342. Power of Four 问题: 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?
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] Power of Two 判断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 这道题让我们判断一个数是否为2的次方数,而且要求时间和空间复杂度...
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!!
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...