50 Pow(x, n) Pow(x, n)
Description:
Implement pow(x, n), which calculates x raised to the power n (xn).
Example:
Example 1:
Input: 2.00000, 10
Output: 1024.00000
Example 2:
Input: 2.10000, 3
Output: 9.26100
Example 3:
Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:
-100.0 < x < 100.0
n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]
题目描述:
实现 pow(x, n) ，即计算 x 的 n 次幂函数。
示例 :
示例 1:
输入: 2.00000, 10
输出: 1024.00000
示例 2:
输入: 2.10000, 3
输出: 9.26100
示例 3:
输入: 2.00000, -2
输出: 0.25000
解释: 2-2 = 1/22 = 1/4 = 0.25
说明:
-100.0 < x < 100.0
n 是 32 位有符号整数，其数值范围是 [−2^31, 2^31 − 1] 。
思路:
快速幂算法
每次将 n减半, 如果是奇数, 乘 x, 否则 x自乘
时间复杂度O(lgn), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
double myPow(double x, int n)
{
double result = 1.0;
for (int i = n; i != 0; i /= 2)
{
if (i & 1) result *= x;
x *= x;
}
return n < 0 ? 1 / result : result;
}
};
Java:
class Solution {
public double myPow(double x, int n) {
double result = 1.0;
for (int i = n; i != 0; i /= 2) {
if ((i & 1) != 0) result *= x;
x *= x;
}
return n < 0 ? 1 / result : result;
}
}
Python:
class Solution:
def myPow(self, x: float, n: int) -> float:
return x ** n
