leetcode 172. Factorial Trailing Zeroes
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Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
思路:要求n!中有多少个0
10的因子有2,5,5即解题关键(因为有5必有2)
2*5=10有一个0
10也有一个0
因此在一个[1,10]区间内有2个0
又因为,25有两个5,125有三个5,它们的倍数会比多一个0
class Solution(object): def trailingZeroes(self, n): """ :type n: int :rtype: int """ count=0 tmp=5 while(n/tmp): count=count+n/tmp tmp=tmp*5 return count
0 0
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