題目描述
給定一個double類型的浮點數base和int類型的整數exponent。求base的exponent次方。
思路:首先先考慮特殊輸入,考慮特殊的輸入值,這裏把0的負數次方定義爲返回值爲0,把0的0次方定位爲1。爲了保證效率,這裏採用 “>>1”作爲除2的方式,因爲2進制移位效率更高,下邊用“exponent & 0x01 == 1”確定奇數偶數也是爲了提升效率。注意每次指數爲奇數是,都要乘一次base,比如求2的7次方,7/2=3,此時會丟掉一次base,需要用奇偶判斷乘上,這個地方在下面的代碼有詳解。
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;//當exponent=1,exponent/2=0 會用到
}
result = powerWithUnsignedInt(base, exponent >> 1);
result *= result;
if ((exponent & 0x1) == 1) {// 每逢奇數都要乘一次base,比如2^7,7/2=3,會丟失一次base值
result *= base;
}
return result;
}
}