[leetcode] 172. Factorial Trailing Zeroes
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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.
这道题是求阶乘结果尾部0的个数,题目难度为Easy。
尾部的0都是由5乘以偶数得来的,也就是说n!中每有一个因子5,尾部就会有一个0,而5的乘方要当做多个5(5本身除外)来处理,因为它有多个因子5,这样问题就转化为统计n!中一共有多少因子5,处理起来就比较简单了,具体代码:
class Solution {public: int trailingZeroes(int n) { int ret = 0; while(n) { n /= 5; ret += n; } return ret; }};
0 0
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