LeetCode(50) Pow(x, n)

处理数值计算问题的两种方法：1、基于分治法的二进制分解的方法；2、基于减二法的二分法。参考Code Ganker。

基于分治法的二进制分解的方法

class Solution {
public:
    double myPow(double x, int n) {


        bool isNegative = false;

        if(n % 2 == 1 && x < 0)

            isNegative = true;

        x = abs(x);

        if(n < 0)  {

            x = 1 / x;
            n = -n;

        }

        double result = 1.0;

        while(n > 0) {

            if(1 == (n & 1))

                result *= x;

            x *= x;
            n = n >> 1;

        }

        return isNegative == false ? result : -result;

    }
};

基于减二法的二分法

class Solution {
public:

    double minusTwo(double x, int n) {

        if(n == 1)

            return x;

        double result = minusTwo(x, n / 2);

        if(1 == n % 2) {

            result = result * result * x;

        }else { 

            result = result * result;

        }

        return result;

    }

    double myPow(double x, int n) {

        if(0 == n) 

            return 1.0;

        bool isNegative = false;

        if(n % 2 == 1 && x < 0)

            isNegative = true;

        x = abs(x);

        if(n < 0)  {

            x = 1 / x;

            if(-2147483648 == n)

                n = 2147483647;

            else

                n = -n;

        }


        return true == isNegative ? -minusTwo(x, n) : minusTwo(x, n);
    }
};