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:首先先考虑特殊输入,考虑特殊的输入值,这里把0的负数次方定义为返回值为0,把0的0次方定位为1。为了保证效率,这里采用 “>>1”作为除2的方式,因为2进制移位效率更高,下边用“exponent & 0x01 == 1”确定奇数偶数也是为了提升效率。注意每次指数为奇数是,都要乘一次base,比如求2的7次方,7/2=3,此时会丢掉一次base,需要用奇偶判断乘上,这个地方在下面的代码有详解。
public class Solution {
public double Power(double base, int exponent) {
boolean minus = false;
double result = 0.0;
if (base == 0) {
if (exponent == 0) {
return 1;
}
return 0.0;
}
if (exponent < 0) {
minus = true;
exponent = 0 - exponent;
}
result = powerWithUnsignedInt(base, exponent);
if ( minus == true) {
return 1 / result;
}else {
return result;
}
}
public double powerWithUnsignedInt(double base, int exponent) {
double result = 0.0;
if (base == 1)
return 1;
if (exponent == 1)
return base;
if (exponent == 0) {
return 1.0;//当exponent=1,exponent/2=0 会用到
}
result = powerWithUnsignedInt(base, exponent >> 1);
result *= result;
if ((exponent & 0x1) == 1) {// 每逢奇数都要乘一次base,比如2^7,7/2=3,会丢失一次base值
result *= base;
}
return result;
}
}