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]

 1 class Solution {
 2 public:
 3     double myPow(double x, long n) {
 4         if (n == 0)return 1;
 5         int _n = n;
 6         if (n < 0)n = -n;
 7         double pow = myPow(x, n / 2), ans;
 8         if (n % 2)
 9             ans = pow * pow*x;
10         else
11             ans=pow * pow;
12         if (_n < 0)
13             return 1 / ans;
14         return ans;
15     }
16 };

有点妙哦,但还是很慢

标签：19.2,Pow,double,Output,ans,Input,pow,LeetCode,Example
来源： https://www.cnblogs.com/yalphait/p/10355471.html