Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
解题思路：
计算n能达到的5的最大次幂，算出在这种情况下能提供的5的个数，然后减去之后递归即可，JAVA实现如下：
static public int trailingZeroes(int n) {
if(n<25)
return n/5;
long five=5;
int count=0;
while(n>=five){
five*=5;
count++;
}
int temp=(int) (n/Math.pow(5, count));
return countSum(count)*temp+trailingZeroes(n-temp*(int)Math.pow(5, count));
}
static public int countSum(int count){
if(count==1)
return 1;
else return countSum(count-1)*5+1;
}
Java for LeetCode 172 Factorial Trailing Zeroes
原文：http://www.cnblogs.com/tonyluis/p/4555912.html