数值的整数次方

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剑指offer第十一题，就是实现java中的库函数pow（double base，int exp），得考虑非法输入以及指数为负的情况

package com.zjy.sword2offer;public class Power {public static double power(double base, int exponent) throws Exception{if(base==0.0 && exponent<0)new Exception("invalid input.");int exp = Math.abs(exponent);double result = 1.0;for(int i=1;i<=exp;i++){result *= base;}if(exponent<0)result = 1/result;return result;}public static void main(String[] args) throws Exception{// TODO Auto-generated method stubdouble res = power(0,-3);System.out.println(res);}}

测试的时候发现我对非法输入设置的异常没有作用，后来发现是因为浮点数有误差，用base==0.0的判断是没法生效的，修改后的代码为

package com.zjy.sword2offer;public class Power {public static double power(double base, int exponent) throws Exception{if(equal(base,0.0) && exponent<0)throw new Exception("invalid input.");int exp = Math.abs(exponent);double result = 1.0;for(int i=1;i<=exp;i++){result *= base;}if(exponent<0)result = 1/result;return result;}public static boolean equal(double val1, double val2){if((val1-val2>-0.000001) && (val1-val2<0.000001))return true;elsereturn false;}public static void main(String[] args) throws Exception {// TODO Auto-generated method stubdouble res = power(0,-3);System.out.println(res);}}

以上算法的复杂度为O(n)，还可以降低，利用一个数的8次方可以由他的4次方相乘，能够将复杂度降低到O(logn)

package com.zjy.sword2offer;public class Power {public static double power(double base, int exponent) throws Exception{if(equal(base,0.0) && exponent<0)throw new Exception("invalid input.");int exp = Math.abs(exponent);double result = 1.0;result = multiply(base,exp);if(exponent<0)result = 1/result;return result;}public static boolean equal(double val1, double val2){if((val1-val2>-0.000001) && (val1-val2<0.000001))return true;elsereturn false;}public static double multiply(double a, int b){if(b==0)return 1;if(b==1)return a;double result = multiply(a,b>>1);result *= result;if((b&1)==1)result *= a;return result;}public static void main(String[] args) throws Exception {// TODO Auto-generated method stubdouble res = power(2,10);System.out.println(res);}}

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