Write an algorithm which computes the number of trailing zeros in n factorial.
11! = 39916800, so the out should be 2
1 class Solution { 2 public long trailingZeros(long n) { 3 long cnt = 0; 4 while(n != 0){ 5 cnt += n / 5; 6 n = n / 5; 7 } 8 return cnt; 9 } 10 }
原文：http://www.cnblogs.com/lz87/p/6931290.html