Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

要求时间复杂度为log型

int trailingZeroes(int n) {
    int ret = 0;
    while(n)
    {
        ret += n/5;
        n /= 5;
    }
    return ret;
}

• 2 * 5 则会出现一个0；2出现的次数比5多；所以只要得到多少个5相乘就可以了