Pow(x, n)

Implement pow(x, n).

可以直接调用api

public class Solution {
    public double pow(double x, int n) {
        return Math.pow(x, n);
    }
}

这道题其实和Divide Two Integers有点像

思路也是一样的，直接用2 ^ n次方递增

需要考虑n为负的情况

public class Solution {
    public double pow(double x, int n) {
        double result = 1;      
        boolean isNeg = false;
        if(n < 0)
            isNeg = true;
        n = Math.abs(n);
        
        while(n > 0){
            long tempCount = 1;
            double tempResult = 1;
            tempResult *= x;
            
            while(tempCount * 2 <= n){
                tempResult *= tempResult;
                tempCount += tempCount;
            }//while
            n -= tempCount;
            result *= tempResult;
        }
        if(isNeg)
            result = 1 / result;
        return result;
    }
}