@ChristopherWu
2016-06-16T21:22:28.000000Z
字数 1810
阅读 1394
leetcode
Given an integer, write a function to determine if it is a power of two.
Integer is a power of two means only one bit of n is '1', for example, 100
is 2^2=4
while 110
is 2^2+2^1=6
.
When n<=0, it can't be power of two as 2^-1=0.5
and because the parameter n is int, we can sure that it has only 32 bit, so we count the number of 1 in 32 bits to check whether only one bit of n is '1' when n is positive.
class Solution {
public:
bool isPowerOfTwo(int n) {
if(n<=0)
return false;
int nums_of_one = 0;
for(int i=0; i<32; ++i) {
nums_of_one += n&1;
if(nums_of_one > 1)
return false;
n >>= 1;
}
return true;
}
};
The result of log2(n) in math must be an interger instead of float when integer is a power of two, so we use log2() function in C++11 and check whether log2(n) is an interger by difference between floor(log2(n))
and ceil(log2(n))
.
For example, n=5, log2(5)=2.19722
, floor(2.19722)=2
, ceil(2.19722)=3
, the difference is 1, so it is not power of two.
class Solution {
public:
bool isPowerOfTwo(int n) {
if(n<=0)
return false;
double tmp = log2(n);
int a = floor(tmp);
int b = ceil(tmp);
if(b-a == 0)
return true;
return false;
}
};
We can also use pow(2, log2(n))
instead of floor(log2(n)) - ceil(log2(n))
.
class Solution {
public:
bool isPowerOfTwo(int n) {
if(!n)
return false;
int t = floor(log2(n));
if(pow(2, t) == n)
return true;
return false;
}
};
I didn't come up with this, thanks for dong.wang.1694's solution.
We can know that power of 2 means only one bit of n is '1', for example, 1
, 10
, 100
etc, so n-1 means the other bits will become 1, e.g. 0
, 01
, 011
. Power of 2 minus 1 means all of its digits will negate.
Therefore, the result of n&(n-1)
must be 0.
class Solution {
public:
bool isPowerOfTwo(int n) {
if(n<=0) return false;
return !(n&(n-1));
}
};