50. Pow(x, n)

https://leetcode.com/problems/powx-n/

Implement pow(x, n), which calculates x raised to the power n (xn).

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 [−231, 231 − 1]

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Intuition

if pow < 0

    pow = Math.abs(pow)

    base = 1.0 / base

Cast power to long to prevent against integer overflow when taking Math.abs

while (pow > 0) {

    pow = pow / 2;

}

Above is log N complexity

To compute power, square the number in every iteration - running calculation

If power is odd, multiply answer with number one extra time

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Time - O(log N)

Space - O(1)

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