首页 > 代码库 > HappyLeetcode39:Factorial Trailing Zeroes
HappyLeetcode39:Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
这个是我的代码，但是超时
class Solution {public: int trailingZeroes(int n) { if (n <= 4) return 0; int count = 0; int multi = 1;//计算所需内容累乘的积 for (int i = 5; i <= n; i += 5) { multi *= i; while (multi % 5 == 0) { multi /= 5; count++; } } return count; }};
网上抄的别人的答案
class Solution {
public:
int trailingZeroes(int n)
{
int c = 0;
for( ; n > 4; c += (n/=5) );
return c;
}
};
class Solution {public: int trailingZeroes(int n) { int c = 0; for( ; n > 4; c += (n/=5) ); return c; }};
HappyLeetcode39:Factorial Trailing Zeroes
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。