# 50. Pow(x, n)
Implement pow(x, n), which calculates x raised to the power n (xn).
Example 1:
Input: 2.00000, 10
Output: 1024.00000
Example 2:
Input: 2.10000, 3
Output: 9.26100
Example 3:
Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:
- -100.0 < x < 100.0
- n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]
# Solution
Approach 1: Recursion.
Approach 2: Iteration.
# Code (Python)
Approach 1:
class Solution:
def myPow(self, x: float, n: int) -> float:
if n == 0:
return 1
if n == 1:
return x
if n == -1:
return 1 / x
if n % 2 == 1:
half = self.myPow(x, n // 2)
return half * half * x
else:
half = self.myPow(x, n // 2)
return half * half
Approach 2:
# Code (C++)
Approach 1:
// Recursion.
// Growth: 1, 2, 4, 8, ...
class Solution {
public:
double myPow(double x, int n) {
if (n == 0) return 1;
if (n < 0) x = 1 / x;
long absLongN = abs(long(n));
double pow = myPow(x, absLongN / 2);
// if (n % 2 == 0)
if (!(n & 1))
pow = pow * pow;
else
pow = pow * pow * x;
return pow;
}
};
Approach 2:
// Iteration.
// Growth: 1, 2, 3, 4, ... [..., 4, 3, 2, 1]
class Solution {
public:
double myPow(double x, int n) {
long absLongN = abs(long(n));
double pow = 1;
int growthRate = 1;
double growthBase = x;
while (absLongN > 0)
{
if (absLongN >= growthRate)
{
absLongN -= growthRate;
pow *= growthBase;
growthRate++;
growthBase *= x;
}
else
{
growthRate--;
growthBase /= x;
}
}
if (n < 0)
pow = (double)1 / pow;
return pow;
}
};