LeetCode刷题笔录 Pow(x,n)

Implement pow(x, n).

这题真简洁。上来就循环，n个x乘一遍，结果超时了。

其实可以用二分法，x^n = x^(n/2) * x^(n/2) * x^(n%2)

public class Solution {
    public double pow(double x, int n) {
        double result = powHelper(x, n);
        if(n < 0)
            result = 1 / result;
        return result;
    }
    
    public double powHelper(double x, int n){
        if(n == 0)
            return 1.0;
        double w = powHelper(x, n / 2);
        if(n % 2 == 0)
            return w * w;
        else
            return w * w * x;
    }
}