"Leetcode" 172. Factorial Trailing Zeroes

Title:

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

Note: Your solution should is in logarithmic time complexity.

Tips:

This problem requires the number of end zeros of n factorial. Because when and only a pair of 2 and 5 appear in factorial, a 0 is present at the end, and the number of 2 is much less than 5, so the number of occurrences of 5 can be calculated. One method is to increment 5 each time and calculate the number of factor 5. There is also a faster way to borrow the base 5 logarithm. You can see the code specifically.

Code:

classSolution { Public:    intTrailingzeroes (intN) {intStep = log (n)/log (5); intCount =0;  for(inti =1; I <= step; ++i) {count+ = (N/pow (5, i)); }        returncount; }};

"Leetcode" 172. Factorial Trailing Zeroes