# 50. Pow(x, n)

## 15 Jan 50. Pow(x, n)

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.
• `-104 <= xn <= 104`

class Solution(object):
def myPow(self, x, n):
“””
:type x: float
:type n: int
:rtype: float
“””
if n == 0:
return 1
if n < 0:
x = 1 / x
n = -n
if n % 2:
return x * self.myPow(x, n – 1)
return self.myPow(x * x, n / 2)
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C++ 代码如下：

class Solution {
public:
double myPow(double x, long long n) {
if (n == 0)
return 1;
if (n == 1)
return x;
if (n < 0)
return 1.0 / myPow(x, -n);
if (n % 2 == 1)
return x * myPow(x, n – 1);
else {
double cur = myPow(x, n / 2);
return cur * cur;
}
}
};
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class Solution(object):
def myPow(self, x, n):
“””
:type x: float
:type n: int
:rtype: float
“””
if n == 0:
return 1
if n < 0:
x = 1 / x
n = -n
ans = 1
res = 1
while n:
if n % 2:
ans *= x
n >>= 1
x *= x
return ans
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class Solution:
def myPow(self, x: float, n: int) -> float:
if n == 0:
return 1

if n < 0:
n = -n
x = 1/x

if n % 2:
return x * self.myPow(x, n-1)
return self.myPow(x*x, n/2)