leetcode 172 Factorial Trailing Zeroes（难易度：Easy）
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Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
代码：
int trailingZeroes(int n) {if(n == 0)return 0;int result = 0;for (long long i = 5; n/i > 0; i *= 5)result += n/i;return result;}原题地址：https://leetcode.com/problems/factorial-trailing-zeroes/
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