50. Pow(x, n)

math

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.

• Either x is not zero or n > 0.

• -104 <= xn <= 104

分析

Instead 每次*x, 把步子指数型提高，每次迈x*x的步子，同时n//=2。奇数时候直接*x

class Solution:
    def myPow(self, x: float, n: int) -> float:
        #扩大底数x  x=>x2 同时缩小n=//2 这样等于每次指数型扩大步子。
        if n < 0:
            n = -n
            x = 1 / x
        
        res = 1
        while n > 0:
            if n%2 == 1:
                res *= x
                n -= 1
            x *= x
            n //= 2
        return res