快速冪Java(leetcode 50 計算 Pow(x, n)

class Solution { public double myPow(double x, int n) { double res = 1; for(int i = n; i != 0; i /= 2) { if((i & 1) == 1) res *= x; x *= x; } return n < 0 ? 1 / res : res; } }