Java for LeetCode 172 Factorial Trailing Zeroes

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;
	}