Pow(x, n) - Leetcode

Implement pow(x, n).

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实现X^n,用课本的语言说就是分治法X^n = X^(n/2)*X^(n/2)*X^(n%2);

当n为偶数时，X^n = X^(n/2)*X^(n/2) ,【n%2 ==0】； 

当n为奇数时，X^n = X^(n/2)*X^(n/2)*X，【n%2 ==1】

考虑一下n为负数的情况。

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时间复杂度O(logn)，空间复杂度O(1)

public class Solution {    public double pow(double x, int n) {        if(n<0)            return 1.0/powHelper(x, -n);        return powHelper(x,n);    }        public double powHelper(double x, int n){        if(n==0)            return 1.0;        double db=powHelper(x,n/2);        if(n%2 == 0)            return db*db;        return db*db*x;    }}