Problem

Write an algorithm which computes the number of trailing zeros in n factorial.

Challenge

11! = 39916800, so the output should be 2

Note

i是5的倍数，先找有多少个5（1个0），然后找多少个25（2个0），补上，然后多少个125（3个0），补上……

Solution

class Solution {    public long trailingZeros(long n) {        long i = 5;        long count = 0;        while (i <= n) {         //n = 125, i = 5, count = 25; 25个5         //i = 25, count += 5（5个25）＝ 30; i = （1个）125, count  += 1 = 31;             count += n / i;            i = i * 5;        }        return count;    }}

LeetCode version

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