问题描述：

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

Note: Your solution should be in logarithmic time complexity.

基本思想

尾0的个数取决于5的个数。

代码：

int trailingZeroes(int n) { //C++
        if(n <= 0)
            return 0;
        int num = 0;
        int size;
        while(n!=0){
            size = n/5;
            num += size;
            n = size;
        }
        return num;
    }

原文：http://blog.csdn.net/chenlei0630/article/details/42271019