Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

思路：实现很简单,关键的分析题意，要求n!末尾的0数，事实上可以求5和2的对数。而又因为出现5必定出现2,所以求5个数即可.

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