```
public boolean isPerfectSquare(int num) {
int i = 1;
while (num > 0) {
num -= i;
i += 2;
}
return num == 0;
}
```

The time complexity is **O(sqrt(n))**, a more efficient one using binary search whose time complexity is **O(log(n))**:

```
public boolean isPerfectSquare(int num) {
int low = 1, high = num;
while (low <= high) {
long mid = (low + high) >>> 1;
if (mid * mid == num) {
return true;
} else if (mid * mid < num) {
low = (int) mid + 1;
} else {
high = (int) mid - 1;
}
}
return false;
}
```

One thing to note is that we have to use **long** for mid to avoid **mid*mid** from overflow. Also, you can use **long** type for **low** and **high** to avoid type casting for mid from long to int.

And a third way is to use **Newton Method** to calculate the square root or num, refer to Newton Method for details.

```
public boolean isPerfectSquare(int num) {
long x = num;
while (x * x > num) {
x = (x + num / x) >> 1;
}
return x * x == num;
}
```