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:
Search
Search-100.0 < x < 100.0-231 <= n <= 231-1-104 <= xn <= 104class Solution: def myPow(self, x: float, n: int) -> float: def helper(x, n): if x == 0: return 0 if n == 0: return 1 res = helper(x, n // 2) res = res * res res = res if n % 2 == 0 else res * x return res res = helper(x, abs(n)) return res if n >= 0 else 1 / res
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