Evaluate a Floating Base to an Integer Power
Company: Google
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Other
### Problem
Implement `myPow(x,n)` to return the floating-point value of `x` raised to the integer exponent `n`. A negative exponent denotes the reciprocal power, and a nonzero base raised to zero is 1.
```cpp
double myPow(double x, int n);
```
Return a numeric value; the decimal formatting in examples is illustrative.
### Examples
```text
Input: x = 3.0, n = -3
Output: 0.037037037037...
```
```text
Input: x = -1.5, n = 4
Output: 5.0625
```
### Constraints
- `-100.0 < x < 100.0`.
- `n` is an integer from -2147483648 through 2147483647.
- Either `x` is nonzero or `n > 0`; zero to a zero or negative exponent is excluded.
- The mathematical result is between -10000 and 10000, inclusive.
Overview: Compute a floating-point power for the full signed 32-bit exponent range, including negative exponents and zero-base constraints.
Community answers
Answer by Coderka14
#include
using namespace std;
class Solution
{
public:
double myPow(double x, int n)
{
long long power = n;
if (power < 0)
{
x = 1.0 / x;
power = -power;
}
double ans = 1.0;
while (power > 0)
{
if (power % 2 == 1)
{
ans *= x;
}
x *= x;
power /= 2;
}
return ans;
}
};