【剑指offer】面试题16：数值的整数次方

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完整代码地址

题目

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

在代码注释里

代码（解法1）

不知道为什么我的第一想法就是拿一个数组来保存值…可是完全没有必要，解法1这个代码浪费了没必要的空间。见解法2。

/** * 题目： * 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： * public double Power(double base, int exp) * 即 base ^ exp * 如果exp为偶数，则拆分成(base ^ (exp/2)) * (base ^ (exp/2)) * 如果exp为奇数，则拆分成(base ^ (exp/2)) * (base ^ (exp/2)) * base *  * 注意：exp可能为负数，负数时先将 exp 转换成绝对值，求出答案 res，最终答案为 1/res  *  * @author peige */public class _16_Power {    public double Power(double base, int exponent) {        if(base == 0)            return 0;        if(exponent == 0)            return 1;        boolean positive = exponent > 0;        exponent = positive ? exponent : -exponent;        double[] power = new double[exponent+1];        power[0] = 1;        power[1] = base;        if(positive)            return Power(base, exponent, power);        else             return 1 / Power(base, exponent, power);    }    private double Power(double base, int exponent, double[] power) {        if(power[exponent] != 0)             return power[exponent];        double tmp = Power(base, exponent/2, power);        if(exponent % 2 == 0)            power[exponent] = tmp * tmp;        else            power[exponent] = tmp * tmp * base;        return power[exponent];    }}

解法2

/** * 题目： * 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： * public double Power(double base, int exp) * 即 base ^ exp * 如果exp为偶数，则拆分成(base ^ (exp/2)) * (base ^ (exp/2)) * 如果exp为奇数，则拆分成(base ^ (exp/2)) * (base ^ (exp/2)) * base *  * 注意：exp可能为负数，负数时先将 exp 转换成绝对值，求出答案 res，最终答案为 1/res  *  * @author peige */public class _16_Power {    public double Power(double base, int exponent) {        if(base == 0)            return 0;        if(exponent == 0)            return 1;        boolean positive = exponent > 0;        exponent = positive ? exponent : -exponent;        if(positive)            return PowerCore(base, exponent);        else             return 1 / PowerCore(base, exponent);    }    private double PowerCore(double base, int exponent) {        if(exponent == 0)            return 1;        if(exponent == 1)            return base;        double tmp = Power(base, exponent / 2);        if(exponent % 2 == 0)            return tmp * tmp;        else            return tmp * tmp * base;    }}

解法3（优化到极致）

public double Power(double base, int exponent) {    if(base == 0)        return 0;    if(exponent == 0)        return 1;    boolean positive = exponent > 0;    exponent = positive ? exponent : -exponent;    if(positive)        return PowerCore(base, exponent);    else         return 1 / PowerCore(base, exponent);}private double PowerCore(double base, int exponent) {    if(exponent == 0)        return 1;    if(exponent == 1)        return base;    //double tmp = Power(base, exponent / 2);    double tmp = Power(base, exponent >> 1);    //if(exponent % 2 == 0)    if((exponent & 1) == 0)        return tmp * tmp;    else        return tmp * tmp * base;}

测试

public static void main(String[] args) {    test1();    test2();}/** * 功能测试 * 1.正数次方 * 2.负数次方 */private static void test1() {    _16_Power power = new _16_Power();    double res = power.Power(1.1, 2);    MyTest.equal(res, 1.21);    res = power.Power(3, 3);    MyTest.equal(res, 27);    res = power.Power(-2, 3);    MyTest.equal(res, -8);    res = power.Power(5, -2);    MyTest.equal(res, 0.04);    res = power.Power(-2, -3);    MyTest.equal(res, -0.125);}/** * 边界测试 * 1.base为0 * 2.exponent为0 */private static void test2() {    _16_Power power = new _16_Power();    double res = power.Power(0, 5);    MyTest.equal(res, 0);    res = power.Power(3, 0);    MyTest.equal(res, 1);}

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