[Leetcode] Factorial Trailing Zeroes

Title: (Math)

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

Note:your solution should is in logarithmic time complexity.

Exercises

Let's start by listing a few examples to find out the rules.

and found

5! There's a 0,

10! There are two of 0,

15! There are three of 0,

20! There are four of 0,

25! There are six of 0,.

It can be seen that when n is greater than or equal to 5 it can produce a suffix of 0, because 5*2=10. and 2 is certainly not missing because as long as there is an even number 2.

So we just need to count n!. Contains 5 of the number can be.

At first, I thought it would be as long as N/5. Later found, for example, 25, contains two 5,25=5*5. So it is found that the total number of 5 equals the number after N/5 divided by 5 until it is less than 5.

 Public classSolution { Public intTrailingzeroes (intN) {if(n<4)          return 0; intres=0;  while(n>=5) {res+=n/5; N=n/5; }        returnRes; }}

[Leetcode] Factorial Trailing Zeroes