傻逼题目。不用多想。
public class Solution {
public boolean isPowerOfTwo(int n) {
if (n <= 0)
return false;
boolean ret = false;
for (int i = 0; i < 32; i++) {
if ((n & 0x0001) == 1) {
if (!ret)
ret = true;
else {
ret = false;
break;
}
}
n = (n >> 1);
}
return ret;
}
}
**
总结： bit manipulation
刷了两道简单题，神清气爽！
**
Anyway, Good luck, Richardo!
My code:
public class Solution {
public boolean isPowerOfTwo(int n) {
if (n < 0) {
return false;
}
int c = 0;
int base = 1;
for (int i = 0; i < 32; i++) {
if ((n & base) != 0) {
c++;
if (c > 1) {
return false;
}
}
base <<= 1;
}
return c == 1;
}
}
这道题目并不难。
Anyway, Good luck, Richardo! -- 10/12/2016
My code:
public class Solution {
public boolean isPowerOfTwo(int n) {
return n > 0 && Math.pow(2, (int) (Math.log(Integer.MAX_VALUE) / Math.log(2))) % n == 0;
}
}
受 power of three 启发。
Anyway, Good luck, Richardo! -- 10/14/2016
