大数阶乘，代码需要优化，求add方法和multiplyDigit（）三目运算符优化方法

[bjh920125]

class Huge
{
private int[] digits;

public Huge(int nDigits)
{
digits = new int[nDigits];
}

private Huge add(Huge n2)
{
Huge result = new Huge(digits.length);

int carry = 0;
for (int k = 0; k < digits.length; k++)
{
int sum = carry + digits[k] + n2.digits[k];
result.digits[k] = sum % 10;
carry = sum / 10;
}
return result;
}

private Huge multiplyDigit(int digit)
{
Huge result = new Huge(digits.length);

int carry = 0;
for (int k = 0; k < digits.length; k++)
{
int prod = carry + digit * digits[k];
result.digits[k] = prod % 10;
carry = prod / 10;
}
return result;
}

private Huge multiply(int number)
{
int weight = 0;
Huge result = new Huge(digits.length);

while (number > 0)
{
int d = number % 10;
number /= 10;
Huge n = multiplyDigit(d);
n.shift(weight++);
result = result.add(n);
}
return result;
}

private void output()
{
int k = digits.length - 1;
while (digits[k--] == 0) ;

for (k = k + 1 ; k >= 0; k--)
System.out.printf("%d", digits[k]);
System.out.println();
}

private void shift(int weight)
{
for (int k = digits.length - 1; k >= 0; k--)
digits[k] = (k >= weight)? digits[k - weight] : 0;
}

public static void main(String[] args)
{
int N = Integer.parseInt(args[0]);
// calculate num-of-digits required for storing factorial(n)
int nbits = 0;
for (int k = 1; k <= N; k++)
nbits += (int)Math.ceil(Math.log10(k));

Huge result = new Huge(nbits);
result.digits[0] = 1;
for (int k = 2; k <= Integer.parseInt(args[0]); k++)
result = result.multiply(k);
result.output();
}
}

[shine333]

来拜神吧，解法及原理内详 http://bbs.csdn.net/topics/340211225#post-342654077 当年，他的回复瞬间将我自认为很不错的解法（基于多线程）给比下去了，而且是瞬秒

[bjh920125]

引用 1 楼 shine333 的回复:

来拜神吧，解法及原理内详 http://bbs.csdn.net/topics/340211225#post-342654077 当年，他的回复瞬间将我自认为很不错的解法（基于多线程）给比下去了，而且是瞬秒

给跪了