数值的整数次方（代码的完整性）

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题目描述：给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。

思路一：简单快速幂法

public class Solution {    public double Power(double base, int exponent) {        int n = 0;        double res = 1;        double current = base;        if (exponent > 0)        {            n = exponent;        }        else if (exponent < 0)        {            if (base == 0) throw new RuntimeException("分母不为0");            n = -exponent;        }        else        {            if (base == 0) return 1;            n = 0;        }        while (n != 0)        {            if ((n & 1) == 1) res = res * current;            current = current * current;            n = n >> 1;        }        return exponent >= 0 ? res : 1/res;    }}

思路二：递归

n为偶数，a^n = a^n/2 * a^n/2；

n为奇数，a^n=（a^（n-1）/2）*（a^（n-1/2））* a
时间复杂度O（logn）

public class Solution {    public double Power(double base, int exponent) {        int n = Math.abs(exponent);        if (n == 0) return 1;        if (n == 1) return base;        double res = Power(base, (n >> 1));        res *= res;        if ((n & 1) == 1) res *= base;        return exponent > 0 ? res : 1/res;    }}

思路三：传统解法

传统公式求解时间复杂度O(n)

public class Solution {    public double Power(double base, int exponent) {        if (base == 0 && exponent == 0) return 1;        double res = 1;        for (int i = 0; i < Math.abs(exponent); i++)        {            res *= base;        }        return exponent > 0 ? res : 1/res;     }}

思路四：利用Math类的pow方法

public class Solution {    public double Power(double base, int exponent) {        if (base == 0.0 && exponent == 0) return 1.0;        return Math.pow(base, exponent);    }}

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