Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
cracking interview原题，2*5可以构成一个10，而5的个数少于2的个数，因此只需找出多少的5即可。
public class Solution {
public int trailingZeroes(int n) {
int num = 0;
while (n >= 5 ) {
num += n / 5;
n = n / 5;
}
return num;
}
}
172. Factorial Trailing Zeroes
原文：http://www.cnblogs.com/shini/p/4421248.html