Factorial Trailing Zeroes

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：有多少个2与5相乘，就有多少个0，由于2远多于5，就是算有多少个5。

class Solution {
public:
    int trailingZeroes(int n) {
        int len = 0;
        while(n >= 5)
        {
            len = len + n/5;
            n = n/5;
        }
        return len;
    }
};