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。
解题思路
- xn = (x * x)n/2 n%2 = 0;
- xn = x * (x2)n/2 n%2 = 1;
代码
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public class {
public double Power(double base, int exponent) {
if (exponent == 0){
return 1;
}
if (exponent == 1){
return base;
}
boolean isNegative = false;
if (exponent < 0) {
exponent = -exponent;
isNegative = true;
}
double pow = Power(base * base, exponent / 2);
if (exponent % 2 != 0){
pow = pow * base;
}
return isNegative ? 1 / pow : pow;
}
}
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