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]
可能是我想多了,因为int的负数范围比正数范围多一,转化为正整数要考虑他的一个范围
class Solution { public: double myPow(double x, int n) { long long m=n; if(n<0){return pow(1/x, 0-m);} return pow(x, n); } double pow(double x, long long n) { if(0==n){return 1.0;} double a=0.0; a = pow(x, n/2); // cout<<a<<" "<<n<<endl; if(n%2){ return a*a*x; }else{ return a*a; } } };
感觉还是不太好,找到一篇文章写的很好
https://blog.csdn.net/gao__xue/article/details/80021207
原文地址:https://www.cnblogs.com/mangmangbiluo/p/9895424.html
时间: 2024-11-05 22:44:10