[LeetCode]50. Pow(x, n)

https://leetcode.com/problems/powx-n/?tab=Description

生成一个pow函数----求幂指数

递归，pow(x, n) == (n % 2) * x * pow(x * x, n / 2)

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