Pow(x, n) Leetcode


#1

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 [−231, 231 − 1]

#2

Solution uses recursion to calculate at double the speed.

func myPow(x float64, n int) float64 {
    if n==0{
        return 1.0
    }
    
    if n<0{
        return 1/myPow(x, -n)
    }
    
    if n%2==1{
        return x * myPow(x, n-1)
    }
    
    return myPow(x*x, n/2)
}

#3

The below solution takes care of scenarios when n is Integer MIN_VALUE.

class Solution {
    public double myPow(double x, int n) {
        if(n == 0)
            return 1;
        if(n == Integer.MIN_VALUE){
            return myPow(x*x, n/2);
        }
        if(n < 0){
            x = 1/x;
            n = -n;
        }
        if(n%2 == 1) 
            return myPow(x*x, n/2)*x;
        else
            return myPow(x*x, n/2);
    }
}