https://leetcode.com/problems/powx-n/description/
一道无聊的二分。注意 n 为 Integer.MIN_VALUE的情况，-n 为 0.
class Solution {
public double myPow(double x, int n) {
if (n == 0) {
return 1;
}
if (n == Integer.MIN_VALUE) {
return myPow(x, -1) * myPow(x, n + 1);
}
if (n < 0) {
x = 1 / x;
n = -n;
}
return (n % 2 == 0) ? myPow(x*x, n/2) : x * myPow(x*x, n/2);
}
}
