【Python3解leetcode Factorial Trailing Zeroes】的更多相关文章

Python3解leetcode Factorial Trailing Zeroes

问题描述: Given an integer n, return the number of trailing zeroes in n!. Example 1: Input: 3 Output: 0 Explanation: 3! = 6, no trailing zero. Example 2: Input: 5 Output: 1 Explanation: 5! = 120, one trailing zero. Note: Your solution should be in logari…

LeetCode Factorial Trailing Zeroes Python

Factorial Trailing Zeroes Given an integer n, return the number of trailing zeroes in n!. 题目意思: n求阶乘以后,其中有多少个数字是以0结尾的. 方法一: class Solution: # @return an integer def trailingZeroes(self, n): res = 0 if n < 5: return 0 else: return n/5+ self.trailingZe…

[LeetCode] Factorial Trailing Zeroes 求阶乘末尾零的个数

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. Credits:Special thanks to @ts for adding this problem and creating all test cases. 这道题并没有什么难度,是让求一个数的阶乘末尾0的个数,也就是要找乘数中10的个数,…

LeetCode Factorial Trailing Zeroes

原题链接在这里:https://leetcode.com/problems/factorial-trailing-zeroes/ 求factorial后结尾有多少个0,就是求有多少个2和5的配对. 但是2比5多了很多,所以就是求5得个数.但是有的5是叠加起来的比如 25,125是5的幂数,所以就要降幂. e.g. n = 100, n/5 =20, n/25= 4, n/125=0,所以加起来就有24个0. O(logn)解法: 一个更聪明的解法是:考虑n!的质数因子.后缀0总是由质因子2和质因…

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: int trailingZeroes(int n) { >)?trailingZeroes(n/)+n/:; } 首先这是LeetCode中时间复杂度为O(logn)的解法. 可以简单的知道,阶乘结果中后缀0的个数取决于n!中因数5的个数…

[LeetCode] Factorial Trailing Zeroes 阶乘末尾0

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. Credits:Special thanks to @ts for adding this problem and creating all test cases. Hide Tags Math   这题应该是2014年年底修改该过测试样本,之前的…

LeetCode Factorial Trailing Zeroes (阶乘后缀零)

题意:如标题 思路:其他文章已经写过,参考其他. class Solution { public: int trailingZeroes(int n) { <? n/: n/+trailingZeroes(n/); } }; AC代码…

LeetCode 172. 阶乘后的零(Factorial Trailing Zeroes)

172. 阶乘后的零 172. Factorial Trailing Zeroes 题目描述 给定一个整数 n,返回 n! 结果尾数中零的数量. LeetCode172. Factorial Trailing Zeroes 示例 1: 输入: 3 输出: 0 解释: 3! = 6,尾数中没有零. 示例 2: 输入: 5 输出: 1 解释: 5! = 120,尾数中有 1 个零. 说明: 你算法的时间复杂度应为 O(log n). Java 实现 递归 class Solution { publi…

【LeetCode】172. Factorial Trailing Zeroes

Factorial Trailing Zeroes Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. Credits:Special thanks to @ts for adding this problem and creating all test cases. 对n!做质因数分解n!=2x*…

LeetCode Day4——Factorial Trailing Zeroes

/* * Problem 172: Factorial Trailing Zeroes * Given an integer n, return the number of trailing zeroes in n!. * Note: Your solution should be in logarithmic time complexity. */ /* * Solution 1 * 对于每一个数字,累计计算因子10.5.2数字出现的个数,结果等于10出现的个数,加上5和2中出现次数较少的 *…