50. Pow(x, n)

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;
    }
};