This code implements an algorithm to calculate \( x^n \)...

This code implements an algorithm to calculate ( x^n ) (x raised to the power of n) using recursion, part of the "power function" concept. It is part of a typical LeetCode problem (myPow or similar) and solves it efficiently by splitting the computation to reduce the number of multiplications needed.

Here’s a step-by-step explanation of what the code does:

1. Special Cases:

   • If ( x == 1 ) or ( n == 0 ), the function immediately returns 1 because any number raised to the power of 0 is 1, and 1 raised to any power remains 1.

2. Base Case:

   • If ( n == 1 ), the function simply returns ( x ) since any number to the power of 1 is the number itself.

3. Recursive Division for Optimization:

   • If ( n ) is even: The function uses the property ( x^n = (x^{n/2})^2 ). In this case, instead of calculating ( x^n ) directly, it calls itself with ( n/2 ).
   • If ( n ) is odd: The function uses the property ( x^n = x \times x^{n-1} ). In this case, it returns ( x ) multiplied by the result of the recursive call with ( n-1 ).

The goal of the recursive approach is to break the problem into smaller sub-problems, reducing the number of multiplications.

Missing Negative Power Handling

This code does not handle negative powers (i.e., if ( n < 0 )). In general, ( x^n ) for negative ( n ) would be computed as ( 1/x^{-n} ).

Example:

If you call myPow(2, 5), the steps would be as follows:

• ( 2^5 = 2 \times 2^4 )
• ( 2^4 = (2^2)^2 )
• ( 2^2 = (2^1)^2 )
• ( 2^1 = 2 )

The recursive calls compute the result step by step, ultimately returning ( 32 ).