问题描述

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

阶乘末尾0的个数

分析

偶数乘以5末尾才会出现0,计算1-n出现含有5的个数即可。

代码

class Solution {
public:
	int trailingZeroes(int n) {
		int r = 0;
		while (n) {
			r += n / 5;
			n /= 5;
		}
		return r;
	}
};