leetcode快速幂—— 50. Pow(x, n)

class Solution {    public double myPow(double x, int n) {        int flag = n >= 0 ? 1 :  -1;        double res = quickPow(x, Math.abs(n));        if(flag == 1) return res;        else return 1.0 / res;    }    public double quickPow(double x, int n) {        if(n==0) return 1;        int k = n / 2;        double res = quickPow(x, n/2);        if(k*2 == n) {            return res * res;        } else {            return res * res * x;        }    }    public static void main(String[] args) {        Solution solution = new Solution();        System.out.println(solution.myPow(2, -1));    }}