* param n: As desciption * return: An integer, denote the number of trailing zeros in n! */publiclongtrailingZeros(longn){// write your code herereturnn /5==0?0: n /5+ trailingZeros(n /5); } }; C++版 class Solution { public: inttrailingZeroes(int n) { return n ==0?0: ...
publicinttrailingZeroes2(intn){if(n<5)return0;if(n<10)return1;returnn/5+trailingZeroes2(n/5); } 这是使用循环结构的解法。 publicinttrailingZeroes3(intn){intresult =0;while(n>0) { result += n/5; n /=5; }returnresult; } 还有更加疯狂的,一行代码搞定。 publicinttrailingZeroes4(int...
Can you solve this real interview question? Factorial Trailing Zeroes - Given an integer n, return the number of trailing zeroes in n!. Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1. Example 1: Input: n = 3 Output: 0 Explanation: 3! = 6,
https://leetcode.com/problems/factorial-trailing-zeroes/ 题目: Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 思路: 打印结果,然后找规律,发现每隔5个周期0的个数增一,隔5*5再上一轮的基础上再加一,直到5^x>=n。。
[leetcode] 172. Factorial Trailing Zeroes Description Given an integer n, return the number of trailing zeroes in n!. Example 1: AI检测代码解析 Input: 3 Output: 0 Explanation: 3! = 6, no trailing zero. 1. 2. 3. Example 2: AI检测代码解析...
172. Factorial Trailing Zeroeswindliang 互联网行业 开发工程师 来自专栏 · LeetCode刷题 4 人赞同了该文章 题目描述(简单难度) 给定一个数,求出一个数的阶乘末尾有多少个 0。 解法一 之前小红书面试的时候碰到的一道题,没想到又是 leetcode 的原题。这种没有通用解法的题,完全依靠于对题目的...
LeetCode上的原题,讲解请参见我之前的博客Factorial Trailing Zeroes。 解法一: inttrailing_zeros(intn) {intres =0;while(n) { res+= n /5; n/=5; }returnres; } 解法二: inttrailing_zeros(intn) {returnn ==0?0: n /5+ trailing_zeros(n /5); ...
Factorial Trailing Zeroes@LeetCode Factorial Trailing Zeroes 这道题目并不难,只要知道了诀窍其实很容。 阶乘,或者说乘法,为什么会产生0,就是因为有2和5的存在,那么只要对输入n进行分解,有多少个2和有多少5就能知道最后的数中会有多少个0。进一步简化,简单地根据规律就可以知道质因子2的数量肯定是多余质因子5的...
Givenanintegern,returnthenumberoftrailing zeroesinn!. Note: Your solution should beinlogarithmictimecomplexity. 分析 起初我看题目的时候没太注意,还以为就是求n这个数后面的零而已,虽然心想不会这么简单吧……就写了一份代码提交了,结果WA提示我5的话应该返回1,这我就纳闷了,5后面毛的0呐……赶紧看题目……...
= 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has two zeroes at the end. Given an integer k, return the number of non-negative integers x have the property that f(x) = k. Example 1: Input: k = 0 Output: 5 Explanation: 0!, 1!, 2!, 3!, and ...