Power button 50 questions Pow (x, n)

This problem is deducted net force on 50 questions.

Achieve  POW ( x ,  n )  , i.e., a function of n-th power of x.

 Recursive and non-recursive thinking python implementation.

class Solution:
     # recurrent ideas 
    DEF myPow_recursion (Self, X, n-):
         IF n-== 0:                                   # recursive termination condition, n == 0 returns. 1 
            return . 1
         IF n-<0:                                    # n-less than 0, it returns to its reciprocal, -n and its recursive structure consistent 
            return . 1 / self.myPow_recursion (X, - n-)
         IF n-. 1 &:                                    # judgment is odd, the result is x * (x n-1 of power) 
            return X * self.myPow_recursion ( x,. 1-n- )
         the else :                                        # judgment is even, the result is the square x * x n / 2-th power 
            returnself.myPow_recursion (X * X, n / 2 )
     # nonrecursive idea 
    DEF myPow_loop (Self, X, n):
         IF n <0:                                    # If n is less than 0, for the number n becomes positive, the reciprocal value 
            n = - n 
            X =. 1 / X 
        POW =. 1                                      # result of the initial value. 1 
        the while n:                                     # traversal cycle, has been divided by n is 2 
            IF n. 1 &:                                # odd time 
                POW = X *                             # POW is X 
            X X * =                                  # X is multiplied by X 
            n->>. 1 =                                  # n-dividing continuity 2 
        return POW
 IF  the __name__ == ' __main__ ' : 
    Solution = Solution ()
     # (2, -1) (2,0) (2,2 &) (2,5) 
    X, = n-2, -2 
    Result = solution.myPow_recursion (X, n-)
     Print (Result) 
    Result = solution.myPow_loop (X, n-)
     Print (Result)