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:
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
19.2.7 [LeetCode 50] Pow(x, n)
原文：https://www.cnblogs.com/yalphait/p/10355471.html
