标签：
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
思路：
首先末尾0的个数取决于素因数中2和5个数较少者，其次5的个数要远远小于2的个数，因此返回素因数中5的个数即可。有公式[n/5] + [n/25] + [n/125] + ... + [n/5x]
解法：
1 public class Solution 2 { 3 public int trailingZeroes(int n) 4 { 5 int numberOfFive = 0; 6 7 while(n / 5 > 0) 8 { 9 n /= 5; 10 numberOfFive += n; 11 } 12 13 return numberOfFive; 14 } 15 }
LeetCode 172 Factorial Trailing Zeroes
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原文地址：http://www.cnblogs.com/wood-python/p/5838470.html
