实现函数double Power(double base, int exponent),求base的exponent次方。不得使用库函数,同时不需要考虑大数问题。
输入: 2.00000, 10
输出: 1024.00000
思路1:
class Solution:
def myPow(self, x: float, n: int) -> float:
if x == 0:
return 0
if n<0:
x = 1/x
n = -n
if n == 0:
return 1
ans = 1
while n > 1:
if n&1:
ans *= x
n -= 1
else:
x = x * x
n = n >> 1
ans *= x
return ans
思路2:
根据奇数和偶数的乘方性质优化代码。 O(logn),O(1)
彻底优化代码,位运算和位移运算比除法和取余运算更加高效。
def offer16_1(base, exponent):
if exponent == 0:
return 1
if exponent == 1:
return base
if exponent < 0:
exponent = -exponent
if base != 0:
base = 1 / base
else:
base = 0
flag = False
if exponent & 1 == 1:
flag = True
res = base
while exponent > 1:
res *= res
exponent = exponent >> 1
if flag
return res * base
else:
return res
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/shu-zhi-de-zheng-shu-ci-fang-lcof