Title description:
Given a floating-point base of type double and an integer exponent of type int. Find the exponent power of base.
Ensure that base and exponent are not 0 at the same time
Example 1
input
2,3
return value
8.00000
Violent solution:
public class Solution {
public double Power(double base, int exponent) {
//暴力解法：时间复杂度o(n),空间复杂度o(1)
if(base == 0 && exponent == 0){
return 0d;
}
if(exponent < 0){
base = 1/base;
exponent = - exponent;
}
double res = 1;
for(int i = 1; i <= exponent; i++){
res = res*base;
}
return res;
}
}
Non-recursive fast power method
public class Solution {
public double Power(double base, int exponent){
/**
*1.全面考察指数的正负、底数是否为零等情况。
*2.写出指数的二进制表达，例如13表达为二进制1101。
*3.举例:10^1101 = 10^0001*10^0100*10^1000。
*4.通过&1和>>1来逐位读取1101，为1时将该位代表的乘数累乘到最终结果。
*/
if(base == 0 && exponent == 0){
return 0d;
}
if(exponent < 0){
base = 1/base;
exponent = -exponent;
}
double res = 1;
while(exponent != 0){
if((exponent&1) == 1){
//表名这一位有1，可以进行累乘
res = res * base;
}
base = base*base;
exponent>>=1;//右移一位
}
return res;
}
}