19.2.7 [LeetCode 50] Pow(x, n)

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]

class Solution {
public:
    double myPow(double x, long n) {
        if (n == 0)return 1;
        int _n = n;
        if (n < 0)n = -n;
        double pow = myPow(x, n / 2), ans;
        if (n % 2)
            ans = pow * pow*x;
        else
            ans=pow * pow;
        if (_n < 0)
            return 1 / ans;
        return ans;
    }
};

有点妙哦,但还是很慢