LeetCode50 Pow(x, n) DIY指数函数

问题描述：
Implement pow(x, n), which calculates x raised to the power n (xn).

Example 1:

Input: 2.00000, 10
Output: 1024.00000

Example 2:

Input: 2.10000, 3
Output: 9.26100

Example 3:

Input: 2.00000, -2
Output: 0.25000

Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:

-100.0 < x < 100.0
n is a 32-bit signed integer, within the range [−231, 231 − 1]
题源：here；完整实现：here

思路：
使用二分的方法加速累乘过程，代码如下：

class Solution {
public:
    double sqr(double a){ return a*a; }
    double myPow(double x, int n) {
        if (n == 0) return 1;
        int flag = n % 2;
        if (n > 0){
            n = n >> 1;
            return flag ? sqr(myPow(x, n))*x : sqr(myPow(x, n));
        }           
        else{
            unsigned int nn = -n;
            nn = nn >> 1; n = nn;
            return flag ? sqr(myPow(x, -n)) / x : sqr(myPow(x, -n));
        }
    }
};