LeetCode Power of Four
原题链接在这里:https://leetcode.com/problems/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?
题解:
与Power of Two, Power of Three类似。
每次iteration若是不能被4整除即return false. 若可以被4整除,便除以4, 直到结果等于1.
Time Complexity: O(log(num)). Space: O(1).
AC Java:
public class Solution {
public boolean isPowerOfFour(int num) {
if(num<=0){
return false;
}
while(num%4 == 0){
num /= 4;
}
return num==1;
}
}
Follow up 需要no loops/recursion.
可以bit manipulation, 首先判定num是否为2的幂数. 若是2的幂数, num二进制表达首位为1, num-1除首位均为1. num & num-1 应等于 0.
然后判定首位的1是在奇数位置上, eg. 10000 = 16. 所以 num & 0x55555555 应等于num 原值.
Time Complexity: O(1). Space: O(1).
public class Solution {
public boolean isPowerOfFour(int num) {
return num>0 && (num & num-1) == 0 && (num & 0x55555555) == num;
}
}
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