Factorial Trailing Zeroes问题及解法
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问题描述:
Given an integer n, return the number of trailing zeroes in n!.
问题分析:
n的阶乘后面有几个零,可根据n以内5的幂决定,举个例子,n=100时,5的倍数有20个,而这20个中5的倍数有4个,最后这4个中5的倍数有0个,所以n=100是,0的个数是24.
以此类推。
过程详见代码:
class Solution {public: int trailingZeroes(int n) { int res = 0; while(n) { res += n / 5; n = n / 5; }return res; }};
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