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?
Credits:
Special thanks to @yukuairoyfor adding this problem and creating all test cases.
给一个有符号的32位整数，写一个函数检查此数是否为4的次方数。
解法1：位操作
解法2：循环
解法3: 数学函数, 换底公式
Java:
public boolean isPowerOfFour(int num) {
int count0=0;
int count1=0;
while(num>0){
if((num&1)==1){
count1++;
}else{
count0++;
}
num>>=1;
}
return count1==1 && (count0%2==0);
} Java:
public boolean isPowerOfFour(int num) {
while(num>0){
if(num==1){
return true;
}
if(num%4!=0){
return false;
}else{
num=num/4;
}
}
return false;
}
Java:
public boolean isPowerOfFour(int num) {
if(num==0) return false;
int pow = (int) (Math.log(num) / Math.log(4));
if(num==Math.pow(4, pow)){
return true;
}else{
return false;
}
}
Python:
class Solution(object):
def isPowerOfFour(self, num):
"""
:type num: int
:rtype: bool
"""
return num > 0 and (num & (num - 1)) == 0 and \
((num & 0b01010101010101010101010101010101) == num)
Python:
# Time: O(1)
# Space: O(1)
class Solution2(object):
def isPowerOfFour(self, num):
"""
:type num: int
:rtype: bool
"""
while num and not (num & 0b11):
num >>= 2
return (num == 1)
C++:
class Solution {
public:
bool isPowerOfFour(int num) {
while (num && (num % 4 == 0)) {
num /= 4;
}
return num == 1;
}
};
C++:
class Solution {
public:
bool isPowerOfFour(int num) {
return num > 0 && int(log10(num) / log10(4)) - log10(num) / log10(4) == 0;
}
};
C++:
class Solution {
public:
bool isPowerOfFour(int num) {
return num > 0 && !(num & (num - 1)) && (num & 0x55555555) == num;
}
};
C++:
class Solution {
public:
bool isPowerOfFour(int num) {
return num > 0 && !(num & (num - 1)) && (num - 1) % 3 == 0;
}
};
类似题目：
[LeetCode] 231. Power of Two 2的次方数
[LeetCode] 326. Power of Three 3的次方数