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

 

LeetCode上的原题，请参见我之前的博客。

 

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 == 0) return 1;        double half = myPow(x, n / 2);        if (n % 2 == 0) return half * half;        else if (n > 0) 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 == 0) return 0;        if (n == 1) return x;        if (n == -1) return 1 / x;        return myPow(x, n / 2) * myPow(x, n - n / 2);    }};