首页 > 代码库 > 231. Power of Two 342. Power of Four -- 判断是否为2、4的整数次幂
231. Power of Two 342. Power of Four -- 判断是否为2、4的整数次幂
231. Power of Two
Given an integer, write a function to determine if it is a power of two.
class Solution {public: bool isPowerOfTwo(int n) { return n > 0 ? (n & (n-1)) == 0 : false; }};
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?
bool isPowerOfFour(int num) { static int mask = 0b01010101010101010101010101010101; //edge case if (num<=0) return false; // there are multiple bits are 1 if ((num & num-1) != 0) return false; // check which one bit is zero, if the place is 1 or 3 or 5 or 7 or 9..., // then it is the power of 4 if ((num & mask) != 0) return true; return false; }
231. Power of Two 342. Power of Four -- 判断是否为2、4的整数次幂
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