50. Pow(x, n)

发布时间：2020-12-14 04:25:48 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]

千万不要用Math.pow(x,n)? 不然面试官会以为你以为他是傻逼

也不能用傻子办法一个一个乘，所以来用每次n/2，然后两项相乘得到暂时的结果，通过recursion得到最终解

class Solution {
    public double myPow(double x,int n) {
       if(n < 0){
           return 1/power(x,-n);
       }
        else return power(x,n);
    }
    public double power(double x,int n){
        if(n==0) return 1;
        double half = power(x,n/2);
        if(n % 2==0) return half * half;
        else return x*half*half;
    }
}

简化版

class Solution {
public:
    double myPow(double x,int n) {
        if (n == 0) return 1;
        double half = myPow(x,n / 2);
        if (n % 2 == 0) return half * half;
        if (n > 0) return half * half * x;
        return half * half / x;
    }
};

（编辑：李大同）

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