codewars--js--Large Factorials--階乘+大數階乘
問題描述:
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
解題思路:
剛開始就是按照尋常情況,直接就用for循環或是遞歸求階乘。然後發現js的Number有位數限制(n數相對較大時,會以科學計數法呈現結果;n數很大時,越界,Infinity)。總之就不能像是題目要求的顯示所有的數字。
參考博客:https://www.cnblogs.com/h5course/p/7566812.html
得出大數相乘,可以用數組來存儲每一位數字,基本求解方法可以類比於小學數學乘法計算。(當24*5時,先用4*5得20,則個位數為0,進位為2;再用2*5+2得12,則十位為2,進位為1,。最後為[0,2,1]。數組倒置後即為乘法結果。)
我的答案:
function factorial(n){ // Add some code if(n<0){return null;} if(n==0 ||n==1){return "1";} let result=[1]; //result數組存儲當前階乘結果for(let num=2;num<=n;num++){ for(let i=0,plus=0 ; i<result.length || plus!=0 ; i++){ let count=(i<result.length)?(num*result[i]+plus):plus; //若當前i小於result所存數字的位數,則分別*num+plus;若等於,則直接進位。 result[i]=count%10; //個位、十位、百位……上的數字,存放在數組result中 plus=(count-result[i])/10; } } return result.reverse().join(""); //將數組result倒序後,即為最後的階乘結果 }
優秀答案:
function factorial(n) { var res = [1]; for (var i = 2; i <= n; ++i) { var c = 0; //c代表進位 for (var j = 0; j < res.length || c !== 0; ++j) { c += (res[j] || 0) * i; res[j] = c % 10; //分別求出個位、十位、百位……的數 c = Math.floor(c / 10); } } return res.reverse().join(""); }
另外發現直接用python做,就沒有這個問題出現。(用遞歸方法)
def fun(n): if n<0: return null elif n==0 or n==1: return "1" else: return n*fun(n-1)
用for循環
def fun(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 return sum
哈哈哈!
codewars--js--Large Factorials--階乘+大數階乘