19.2.7 [LeetCode 50] Pow(x, n)

发布时间：2020-12-14 04:25:44 所属栏目：大数据来源：网络整理

导读：Implement? pow( x ,? n ) ,which calculates? x ?raised to the power? n ?(xn). Example 1: Input: 2.00000,10Output: 1024.00000 Example 2: Input: 2.10000,3Output: 9.26100 Example 3: Input: 2.00000,-2Output: 0.25000Explanation: 2-2 = 1/22 = 1/4

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 };

View Code

有点妙哦,但还是很慢

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!