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;
    }
};