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。
解体思路1
使用普通迭代求解幂次方
代码实现:(O(N))
public class Solution {
public double Power(double base, int exponent) {
boolean isNegative = false;
double result = 1;
if(exponent == 0) {
return 1;
} else if(exponent < 0){
exponent = - exponent;
isNegative = true;
}
for(int i=1; i<=exponent; i++) {
result *= base;
}
return isNegative ? 1/result : result;
}
}
解体思路2
代码实现：(O(logN))
public class Solution {
public double Power(double base, int exponent) {
boolean isNegative = false;
if(exponent == 0)
return 1;
if(exponent == 1)
return base;
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;
}
}