leetcode 50. Pow (x, n) (Fast Power)

https://leetcode-cn.com/problems/powx-n/

实现 pow(xn) ,即计算 x 的 n 次幂函数。

示例 1:

输入: 2.00000, 10
输出: 1024.00000

示例 2:

输入: 2.10000, 3
输出: 9.26100

示例 3:

输入: 2.00000, -2
输出: 0.25000
解释: 2-2 = 1/22 = 1/4 = 0.25

说明:

  • -100.0 < x < 100.0
  • n 是 32 位有符号整数,其数值范围是 [−231, 231 − 1] 。
class Solution {
public:
    double pow2(double x,long long n){
        if(n==0)return 1.0;
        double a=pow2(x,n/2);
        double ans=a*a;
        if(n%2)ans=ans*x;
        return ans;
    }
    double myPow(double x, long long n) {
        if(n<0){
            n=-n;
            x=1/x;
        }
        return pow2(x,n);
    }
};

 


 
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Origin www.cnblogs.com/wz-archer/p/12524416.html