【leetcode】50. Pow(x, n)

Implement pow(x, n).

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需考虑n=0和n=Integer.MIN_VALUE的情况， 

此处求结果用了二分法。

public class Solution {    public double myPow(double x, int n) {        if (n == 0){            return 1;        }        if (n == Integer.MIN_VALUE){            x = x * x;            n = n / 2;        }        if (n < 0){            n = -n;            x = 1 / x;        }        return ((n % 2) == 0) ? myPow(x * x, n / 2) : x * myPow(x * x, n / 2);    }}