Large Factorials

Large Factorials

题目

In mathematics, the factorial of integer n is written as n!. It is equal to the product of n and every integer preceding it. For example: 5! = 1 x 2 x 3 x 4 x 5 = 120

Your mission is simple: write a function that takes an integer n and returns the value of n!.

You are guaranteed an integer argument. For any values outside the non-negative range, return null, nil or None (return an empty string "" in C and C++). For non-negative numbers a full length number is expected for example, return 25! = "15511210043330985984000000" as a string.

For more on factorials, see http://en.wikipedia.org/wiki/Factorial

NOTES:

• The use of BigInteger or BigNumber functions has been disabled, this requires a complex solution
• I have removed the use of require in the javascript language.

分析

阶乘，问题在于js对数字长度有限制，超长之后就会直接返回infinity。
解决方案就是把数字拆成数组，对每一位单独进行运算，按照逢十进一的规则加到数组的下一项里。最后把数组翻转并组合就是所需要的答案

另外，python3里面没有长度限制，直接计算就可以了。额外附上python3的代码。

答案

• JavaScript

function factorial(n) {
    // 按照位数算，把数字用数组表示
    if (n < 0) return null
    if (n < 2) return "1"
    let result = [1];
    for (let i = 2; i < n + 1; i++) {
        for (let j = 0, plus = 0; j < result.length || plus != 0; j++) {
            let count = (j < result.length) ? (i * result[j] + plus) : plus
            result[j] = count % 10
            plus = (count - result[j]) / 10
        }
    }
    return result.reverse().join("")
}

• python3

def factorial(n):
    sum=1
    if n<0:
        return null
    elif n==0 or n==1:
        return "1"
    else:
        for i in range(1,n+1):
            sum*=i
            print len(sum)
        return sum