LeetCode 50. Pow(x,n)

题目链接：点击这里

先介绍一种精度最高的解法：

$$x^n=e^{nlnx}$$

class Solution {
    
public:
    double myPow(double x, int n) {
    
        int sign = 1;
        if(x<0 && n&1)  sign = -1;
        x = abs(x);
        return sign*exp(n*log(x));
    }
};

利用倍增思想，快速幂的递归写法：

class Solution {
    
public:
    double fastPow(double x, long long n) {
    
        if (n == 0) return 1.0;
        double half = fastPow(x, n/2);
        if(n%2==0)  return half * half;
        else    return half * half * x;
    }

    double myPow(double x, int n) {
    
        long long N = n;
        if (N < 0) {
    
            x = 1 / x;
            N = -N;
        }
        return fastPow(x, N);
    }
};