Given an integer n, return the number of trailing zeroes in n!.
Example 1:
Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.
Example 2:
Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.
Note: Your solution should be in logarithmic time complexity.
其实就是寻找n!表达式分解后5的个数,因为5*2=10,而2在5之前,个数一定比5多。而5的次方如25,125,625等含有多个5,连续除以5以计算这些多余的5。就是n/5+n/25+n/125...即n/5+n/5/5+n/5/5/5,如下的while循环:
class Solution(object): def trailingZeroes(self, n): """ :type n: int :rtype: int """ fiveNums = 0 while n >= 5: fiveNums += n / 5 n = n / 5 return fiveNums