LeetCode:Factorial Trailing Zeroes
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问题描述:
Given an integer n, return the number of trailing zeroes in n!.
计算n!中0的个数。
思路:
对n!做质因数分解n!=2x*3y*5z*...
显然0的个数等于min(x,z),并且min(x,z)==z
证明:
对于阶乘而言,也就是1*2*3*...*n
[n/k]代表1~n中能被k整除的个数
那么很显然
[n/2] > [n/5] (左边是逢2增1,右边是逢5增1)
[n/2^2] > [n/5^2](左边是逢4增1,右边是逢25增1)
……
[n/2^p] > [n/5^p](左边是逢2^p增1,右边是逢5^p增1)
随着幂次p的上升,出现2^p的概率会远大于出现5^p的概率。
因此左边的加和一定大于右边的加和,也就是n!质因数分解中,2的次幂一定大于5的次幂
JAVA代码:
public class Solution { public int trailingZeroes(int n) { if(n < 0) return 0; int cnt = 0; while(n != 0){ cnt += n / 5; n /= 5; } return cnt; }}
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