[LeetCode] 50. Pow(x, n)

Implement pow(x, n).
Thought process:
Shortest problem description! 
Recursively calculate pow(x, n / 2). Base cases:
  1. n = 0, return 1.
  2. n = 1, return x.
  3. n = -1, return 1 / x.

Solution:
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public class Solution {
    public double myPow(double x, int n) {
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return x;
        }
        if (n == -1) {
            return 1 / x;
        }
        
        if (n % 2 == 0) {
            double half = myPow(x, n / 2);
            return half * half;
        }
        if (n < 0) {
            return myPow(x, n / 2) * myPow(x, n / 2 - 1);
        }
        return myPow(x, n / 2) * myPow(x, n / 2 + 1);
    }
}

Time complexity: O(logn).
Location: San Jose, CA, USA

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