Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

方法：求乘积中5的个数。

class Solution {public:    int trailingZeroes(int n) {        long num5 = 0; //int overflow        long base = 5;        while(n/base){           num5 += n/base;          base *= 5;        }        return num5;    }};

切忌不可用int类型定义，容易溢出。
防止溢出可以换种编程方式进行。

class Solution {public:    int trailingZeroes(int n) {        long num5 = 0; //int overflow        while(n){           num5 += n/5;          n/= 5;        }        return num5;    }};