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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.
1 class Solution { 2 public: 3 int trailingZeroes(int n) { 4 int res=0; 5 while(n) 6 { 7 res+=n/5; 8 n/=5; 9 } 10 11 return res; 12 } 13 };
【leetcode】Factorial Trailing Zeros
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原文地址：http://www.cnblogs.com/jawiezhu/p/4499282.html
