Pow(x,n) Problem Solution

Pow(x,n) Problem: Implement pow(x, n), which calculates `x` raised to the power `n` (i.e., `xn`).

Example :

```Input: x = 2.00000, n = 10
Output: 1024.00000```

Table of Contents

Problem Solution In Python

``````class Solution:
def myPow(self, x: float, n: int) -> float:
def pow(x, n):
if n == 0:
return 1

r = pow(x, n//2)
if n % 2 == 0:
return r * r
else:
return r * r * x

if n < 0:
n *= -1
x = 1/x

return pow(x, n)
``````

Problem Solution In Java

``````public double myPow(double x, int n) {
double[] stored = new double[32];
stored[0] = x;
for (int i = 1; i < 32; i++) stored[i] = stored[i-1] * stored[i-1];

int exponent = n < 0 ? n == Integer.MIN_VALUE ? Integer.MIN_VALUE : -n : n;
double ans = 1;
for (int i = 0; i < 32; i++) {
if ((exponent&(1<<i)) != 0) {
ans *= stored[i];
}
}

return n > 0 ? ans : 1/ans;
}
``````

Problem Solution In C++

``````class Solution {
public:
double myPow(double x, int n) {
if (x == 1.0) return x;
uint32_t m = (n < 0 ? -static_cast<int64_t>(n) : n);
double z = 1.0,  w = (n < 0 ? 1.0 / x : x);
while (m != 0) {
if (m & 0x1UL) z *= w;
w *= w;
m >>= 1;
}
return z;
}
};
``````

Problem Solution In C

``````double myPow(double x, int n) {
double res = 1;
long absn;

absn =  (n < 0) ? ((long)n * -1) : n;
while (absn > 0) {
if (absn & 1)
res *= x;
x *= x;
absn >>= 1;
}
res = (n < 0) ? 1/res : res;
return(res);
}
``````

If You Like This Page Then Make Sure To Follow Us on Facebook, G News and Subscribe Our YouTube Channel. We will provide you updates daily.