Factorial Trailing Zeroes.md

#Factorial Trailing Zeroes ##Problem: Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity. ##Idea: factor 5 decides the number of trailing zeroes.

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

Wrong Answer:1808548329, int overflow

class Solution {
public:
    int trailingZeroes(int n) {
        int count = 0;
        int base = 1;
        int boundary=n/5;
        while(base<=boundary)
        {
            base *= 5;
            count += n/base;
        }
        return count;
    }
};

##To Study:

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