172. Factorial Trailing Zeroes

题目

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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思路

计算n的阶乘的最后有几个零，就是计算因数中2和5的最小个数，5的个数肯定比2的少，就是计算因数中5的个数

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代码

class Solution {public:    int trailingZeroes(int n) {        //计算n的阶乘的最后有几个零，就是计算因数中2和5的最小个数，5的个数肯定比2的少，就是计算因数中5的个数        if(n <= 4)        {            return 0;        }        int times = 0;        while(n)        {            times += n/5;            n /= 5;        }        return times;    }};