Factorial Trailing Zeroes (Divide-and-Conquer)

QUESTION

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

Note: Your solution should be in logarithmic time complexity.

FIRST TRY

class Solution {
public:
    int trailingZeroes(int n) {
        int divident;
        int nOf2 = ;
        int nOf5 = ;
        while(n% == )
        {
            nOf2++;
            divident = n/;
        }
        while(n% == )
        {
            nOf5++;
            divident = n/;
        }
        return min(nOf2,nOf5);
    }
};

Result: Time Limit Exceeded

Last executed input: 0

SECOND TRY

考虑n=0的情况

class Solution {
public:
    int trailingZeroes(int n) {
        int divident;
        int nOf2 = ;
        int nOf5 = ;
        for(int i = ; i < n; i++)
        {
            divident = i;
            while(divident% == )
            {
            nOf2++;
            divident /= ;
            }
            divident = i;
            while(divident% == )
            {
            nOf5++;
            divident /= ;
            }
        }
        return min(nOf2,nOf5);
    }
};

Result:  Time Limit Exceeded

Last executed input:1808548329

THIRD TRY

2肯定比5多

要注意的就是25这种，5和5相乘的结果，所以，还要看n/5里面有多少个5，也就相当于看n里面有多少个25，还有125，625...

class Solution {
public:
    int trailingZeroes(int n) {
        if(n==) return ;
        int divident=n;
        int nOf5 = ;

        while(divident!= )
        {
            divident /= ;
            nOf5+=divident;
        }

        return nOf5;
    }
};

Result: Accepted