劍指offer:數值的整數次方
阿新 • • 發佈:2018-12-09
/** * 題目: * 給定一個double型別的浮點數base和int型別的整數exponent。求base的exponent次方。 */ public class P110_IntegerPowerOfNumber { public double IntegerPowerOfNumber(double base, int exponent) { double result = 0; //輸入底數為0 if (base == 0) { return 0; } //輸入指數為0 if (exponent == 0) { return 1; } result = 1; int temp = Math.abs(exponent); for (int i = 1; i <= temp; i++) { result *= base; } if (exponent > 0) { result = result; } if (exponent < 0) { result = 1 / result; } return result; } public static void main(String[] args) { double base = 1; int exponent = 0; P110_IntegerPowerOfNumber test = new P110_IntegerPowerOfNumber(); double result = test.IntegerPowerOfNumber(base, exponent); System.out.print(result); } }