[求助]Java编的求最小公倍数的程序，老是不对。。。调试了无数次了，请达人指教。。

//根据老师的要求，用这种算法求最小公倍数。总是有错

import javax.swing.JOptionPane;

public class LCM {

/** Main methed */

public static void main(String[] args) {

// Prompt user input the two numbers

String oneString = JOptionPane.showInputDialog(null,

"Please input the first number:", "Input",

JOptionPane.INFORMATION_MESSAGE);

String twoString = JOptionPane.showInputDialog(null,

"Please input the second number:", "Input",

JOptionPane.INFORMATION_MESSAGE);

// Convert the two strings into int value

double oneDouble = Double.parseDouble(oneString);

double twoDouble = Double.parseDouble(twoString);

double twoNumbersLcm; // 定义的这个字段用来存放最后的结果,也就是两个数的最小公倍数

twoNumbersLcm = lcm(oneDouble, twoDouble);

JOptionPane.showMessageDialog(null, "The lease common multiple is "

+ twoNumbersLcm, "Answer", JOptionPane.INFORMATION_MESSAGE);

}

// 求最小公倍数

public static double lcm(double one, double two) {

double[][] ones = new double[10][2];

double[][] twos = new double[10][2];

double commonMultiple = 1;

divide(ones, one);

divide(twos, two);

commonMultiple = multiple(ones, twos);

return commonMultiple;

}

// 求因子，并将得到的因子和出现的次数存到二维数组中

public static void divide(double array[][], double dividend) {

int i = 0;

double divisor = 2.0;

do {

while (dividend % divisor == 0) {

array[i][0] = divisor;

array[i][1]++;

dividend = dividend / divisor;

}

i++;

divisor++;

} while (dividend != 1);

}

//

public static double multiple(double arrayOne[][], double arrayTwo[][]) {

int i = 0;

double power = 1;

// 去次数较大的因子相乘，得到最小公倍数

while (i<=arrayOne.length || i<=arrayTwo.length) {

if (arrayOne[i][0] == arrayTwo[i][0]) {

if (arrayOne[i][1] >= arrayTwo[i][1])

power *= Math.pow(arrayOne[i][0], arrayOne[i][1]);

else

power *= Math.pow(arrayTwo[i][0], arrayTwo[i][1]);

} else

power = Math.pow(arrayOne[i][0], arrayOne[i][1])

* Math.pow(arrayTwo[i][0], arrayTwo[i][1]);

i++;

}

if (arrayOne[i][0] != 0)

power *= Math.pow(arrayOne[i][0], arrayOne[i][1]);

else

power *= Math.pow(arrayTwo[i][0], arrayTwo[i][1]);

return power;

}

}