In the Leetcode Pow(x, n) problem solution Implement pow(x, n), which calculates x raised to the power n (i.e., xn).
Example 1:
Input: x = 2.00000, n = 10
Output: 1024.00000
Example 2:
Input: x = 2.10000, n = 3
Output: 9.26100
Example 3:
Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
Constraints:
- -100.0 < x < 100.0
- -231 <= n <= 231-1
- n is an integer.
- -104 <= xn <= 104
Solution in C Programming
double myPow(double x, int n){
int i=0;
double res = x;
double tempx= x;
if(n==0 || x==1){
return 1;
}
if(n<0){
res = 1/res;
tempx = res;
n=-1*n;
}
while(n > 1){
res *= tempx;
n--;
}
return res;
}Solution in C++ Programming
class Solution {
public:
double binaryExp(double x,long long n){
if(n==0) return 1;
double call = binaryExp(x,n/2);
if(n&1) return x * call * call;
else return call * call;
}
double myPow(double x, int num){
long long n = num;
return n < 0 ? 1 / binaryExp(x,-n) : binaryExp(x,n);
}
};Solution in Java Programming
import java.lang.Math;
class Solution {
public double myPow(double x, int n) {
return (double)Math.pow(x,n);
}
}Solution in Python Programming
class Solution:
def myPow(self, x, n):
# store base^exp tuples for each
stored = [(1,x)] # (exp,base^exp) lookup table of past calculations
if x == 0: return 0.0
if n == 0:
return 1
elif n > 0:
return power(stored, x,n-1)
else:
return 1/(power(stored, x,-(n+1)))
def power(stored, result, exp):
if exp == 0:
return result
stored_exp, stored_result = find_largest_exp(stored,exp)
# update table
largest_stored_exp = stored[-1][0]
if largest_stored_exp < exp:
stored.append((stored_exp*2,result*stored_result))
return power(stored, result*stored_result, exp-stored_exp)
def find_largest_exp(stored,exp):
for stored_exp, stored_result in stored[::-1]:
if exp >= stored_exp:
return (stored_exp, stored_result)Solution in C# Programming
public class Solution {
public double MyPow(double x, int n)
{
if(n < 0)
{
n *= -1;
x = 1/x;
}
return Half(x,n);
}
private double Half(double x, int n)
{
if(n == 0)
return 1.0;
double half = Half(x, n/2);
if(n%2 == 0)
return half * half;
else
return half*half*x;
}
}Also read,