[LintCode] Pow(x, n) 求x的n次方

Implement pow(x, n).

Notice

You don't need to care about the precision of your answer, it's acceptable if the expected answer and your answer 's difference is smaller than 1e-3.

Have you met this question in a real interview?

Yes

Example

Pow(2.1, 3) = 9.261
Pow(0, 1) = 0
Pow(1, 0) = 1

LeetCode上的原题，请参见我之前的博客Pow(x, n)。

解法一：

class Solution {
public:
    /**
     * @param x the base number
     * @param n the power number
     * @return the result
     */
    double myPow(double x, int n) {
        if (n == ) return ;
        double half = myPow(x, n / );
        if (n %  == ) return half * half;
        else if (n > ) return half * half * x;
        else return half * half / x;
    }
};

解法二：

class Solution {
public:
    /**
     * @param x the base number
     * @param n the power number
     * @return the result
     */
    double myPow(double x, int n) {
        if (n == ) return ;
        if (n == ) return x;
        if (n == -) return  / x;
        return myPow(x, n / ) * myPow(x, n - n / );
    }
};