Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
1 int trailingZeroes(int n) {
2 if(n < 5)
3 return 0;
4 int k = 0;
5 while(n)
6 {
7 n = n / 5;
8 k += n;
9 }
10 return k;
11 }
轉:http://www.jianshu.com/p/211618afc695