Leetcode 172. Factorial Trailing Zeroes

172. Factorial Trailing Zeroes

Total Accepted: 60237 Total Submissions: 182176 Difficulty: Easy

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

Note: Your solution should be in logarithmic time complexity.

思路：

0由2,5相乘得出。5的个数一定小于2。所以只用判断5的个数。需要注意的是5,10,15,20都只提供一个5，但是25提供了两个5！

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