Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
public class Solution { public int trailingZeroes(int n) { // math : n * n-1 * n-2...1 = 2 ^ x * 5 ^ y depend on y int cnt = 0; while(n > 0){ cnt += n/5; n = n/5; //26 or 126 } return cnt; } }
172. Factorial Trailing Zeroes
原文：http://www.cnblogs.com/joannacode/p/6100251.html