算法练习(27) leetcode 50. Pow(x, n)

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

class Solution { public double erfen(double x, long N) { if (N == 0) { return 1.0; } double y = erfen(x, N / 2); return N % 2 == 0 ? y * y : y * y * x; } public double myPow(double x, int n) { long N = n; return N >= 0 ? erfen(x, N) : 1.0 / erfen(x, -N); } }