Java for Leetcode 172 factorial Trailing Zeroes

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

Note: Your solution should is in logarithmic time complexity.

Problem Solving Ideas:

Calculates the maximum power of 5 that can be reached by N, calculates the number of 5 that can be provided in this case, and then subtracts the recursion, and the Java implementation is as follows:

static public int trailingzeroes (int n) {if (n<25) return N/5;long five=5;int count=0;while (n>=five) {Five*=5;count ++;} int temp= (int) (N/MATH.POW (5, Count)), return Countsum (count) *temp+trailingzeroes (n-temp* (int) Math.pow (5, count));} static public int countsum (int count) {if (count==1) return 1;else return Countsum (count-1) *5+1;}

Java for Leetcode 172 factorial Trailing Zeroes