关于[LeetCode]Factorial Trailing Zeroes O(logn)解法的理解

题目描述：

Given an integer n, return the number of trailing zeroes in n!.

题目大意：

给定一个整数n，返回n!（n的阶乘）结果中后缀0的个数（如5!=120，则后缀中0的个数为1）。

解题思路：

1 int trailingZeroes(int n) {
2    return (n/5>0)?trailingZeroes(n/5)+n/5:0;
3 }

首先这是LeetCode中时间复杂度为O(logn)的解法。

未完待续。

时间： 2024-11-08 01:12:06

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