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QUESTION
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
FIRST TRY
class Solution { public: int trailingZeroes(int n) { int divident; int nOf2 = 0; int nOf5 = 0; while(n%2 == 0) { nOf2++; divident = n/2; } while(n%5 == 0) { nOf5++; divident = n/5; } return min(nOf2,nOf5); } };
Result: Time Limit Exceeded
Last executed input: 0
SECOND TRY
考虑n=0的情况
class Solution { public: int trailingZeroes(int n) { int divident; int nOf2 = 0; int nOf5 = 0; for(int i = 1; i < n; i++) { divident = i; while(divident%2 == 0) { nOf2++; divident /= 2; } divident = i; while(divident%5 == 0) { nOf5++; divident /= 5; } } return min(nOf2,nOf5); } };
Result: Time Limit Exceeded
Last executed input:1808548329
THIRD TRY
2肯定比5多
要注意的就是25这种，5和5相乘的结果，所以，还要看n/5里面有多少个5，也就相当于看n里面有多少个25，还有125，625...
class Solution { public: int trailingZeroes(int n) { if(n==0) return 0; int divident=n; int nOf5 = 0; while(divident!= 0) { divident /= 5; nOf5+=divident; } return nOf5; } };
Result: Accepted
Factorial Trailing Zeroes (divide and conquer)
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原文地址：http://www.cnblogs.com/qionglouyuyu/p/4199843.html
