快速幂

Java

class Solution {
    public double myPow(double x, int n) {
        if (n == 0) return 1.0;
        long m = (long)n; // 这里因为Integer.MIN_VALUE,INT_MIN，前面加负号还是本身，所以用long存
        return m > 0 ? qmi(x, m) : 1 / qmi(x, -m);
    }
    public double qmi(double x, long n) {
        double res = 1;
        while (n > 0) {
            if ((n & 1) == 1) res *= x;
            n >>= 1;
            x = x * x;
        }
        return res;
    }
}