Python Leetcode 50. Pow(x, n)

【5分钟力扣】50. 用python3实现Pow(x,n)幂次运算 用5分钟时间学习一下python幂次运算,了解和掌握二分法优化时间复杂度 阅读详情

Python Leetcode 50. Pow(x, n)

在这里插入图片描述

# 递归
class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        def quickmul(N):
            if N ==0:
                return 1.0
            y = quickmul(N/2)
            if N%2 == 0:
                return y*y
            else:
                return y*y*x
        
        return quickmul(n) if n>0 else 1/quickmul(-n)
# 非递归
class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        def quickmul(N):
            ans = 1.0
            x_con = x
            while N>0:
                if N%2 == 1:
                    ans *= x_con
                x_con *= x_con
                N //=2
            return ans

        return quickmul(n) if n>0 else 1.0/quickmul(-n)
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