LeetCode #50 Pow(x, n) Solution & Explanation

LeetCode Problem

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

Solution

In our first try, we multiply by iteration ➡ multiply n times

C# Solution

❌First try ➡ Time Limit Exceeded Error

public class Solution {
    public double MyPow(double x, int n) {
        if(x == 0)
        {
            return 0.0;
        }
        else if (n == 0)
        {
            return 1.0; 
        }
        
        double res = 1;
        bool neg = n < 0;
        if(neg)
        {
            n*=-1;
        }
        
        for(int i = 0;i<n;i++)
        {
            res*=x;
        }
        
        //if x is a negative number
        if(neg)
        {
            res = 1/res;
        }
        
        return res;
    }
}

⭐Solution1

In this solution, as example 2⁸ can write as 2*2*2*2*2*2*2*2

Also can write as 2² * 2² * 2² * 2² or 4² * 4²

So if we can get 4² ,we can simplify the multiplication times.

public class Solution {
    public double MyPow(double x, int n) {
        if(x == 0)
        {
            return 0.0;
        }
        
        double res = cal(x, n);
        return n>0 ? res : 1/res;
    }
    
    
    private double cal(double x, int n)
    {
	//Console.WriteLine("x:"+x+",n="+n);     //*1
        if (n == 0)
        {
            return 1.0; 
        }
            
        double res = cal(x*x,n/2);
        return n%2 == 0 ? res : res*x;
    }
}

you can see the log with the code at *1

Java Solution

Solution1

class Solution {
    public double myPow(double x, int n) {
        if(x == 0)
        {
            return 0.0;
        }
        
        double res = cal(x, n);
        return n>0 ? res : 1/res;
    }
    
    private double cal(double x, int n)
    {
        if (n == 0)
        {
            return 1.0; 
        }
            
        double res = cal(x*x,n/2);
        return n%2 == 0 ? res : res*x;
    }
}

Python3 Solution

Solution1

def cal(x: float, n: int) -> float:
    if n == 0:
        return 1.0 

    res = cal(x*x,(int)(n/2))
    return res if n%2 == 0 else res*x

class Solution:    
    def myPow(self, x: float, n: int) -> float:
        if x == 0:
            return 0
        res = cal(x, n)
        return res if n>0 else 1/res

JavaScript Solution

Solution1

/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function(x, n) {
    if(x == 0)
    {
        return 0.0;
    }

    res = cal(x, n);
    return n>0 ? res : 1/res;
};

function  cal(x, n)
{
    if (n == 0)
    {
        return 1.0; 
    }

    res = cal(x*x,parseInt(n/2));
    return n%2 == 0 ? res : res*x;
}

Conclusion

🧡If my solution helps, that is my honor!

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✅If you got any problem about the explanation or you need other programming language solution, please feel free to let me know (either leave a comment or contact me by Messenger is ok!)

Cover Photo : Photo by Kamran Chaudhry on Unsplash

The problem link : Pow(x, n) – LeetCode

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