LeetCode Top Interview 150

50. Pow(x, n)

Medium

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
• -104 <= xn <= 104

Solution

class Solution:
    def myPow(self, x: float, n: int) -> float:
        nn = n
        res = 1.0
        if n < 0:
            nn = -nn
        while nn > 0:
            if nn % 2 == 1:
                nn -= 1
                res *= x
            else:
                x *= x
                nn //= 2
        if n < 0:
            return 1.0 / res
        return res

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