LeetCode 172. Factorial Trailing Zeroes

172. Factorial Trailing Zeroes
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Total Accepted: 55895 Total Submissions: 171950 Difficulty: Easy
Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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题目大意：在logn的时间内 找到n！末尾有几个零

//5*2 = 10出现一个0，n*(n-1)*(n-2)...1当中能被5整除的数少于能被2整除的数//所以能被5整除的5的个数就是0的个数//比如25！,25 = 5 * 5有两个5，20，15，10，5各含一个5，这六个5分别和2结合相乘就能得到末尾6个0//所以只要count每个因子中5的个数就行class Solution {public:    int trailingZeroes(int n) {        int cnt = 0;        while(n != 0) {            cnt = cnt + n / 5;            n = n / 5;        }        return cnt;    }};