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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 int trailingZeroes(int n) { 2 if(n < 5) 3 return 0; 4 int k = 0; 5 while(n) 6 { 7 n = n / 5; 8 k += n; 9 } 10 return k; 11 }
转：http://www.jianshu.com/p/211618afc695
172. Factorial Trailing Zeroes
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原文地址：http://www.cnblogs.com/boluo007/p/5521642.html
