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;
	}
};