【Factorial Trailing Zeroes——数学类】的更多相关文章

Factorial Trailing Zeroes——数学类

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 转自:http://blog.csdn.net/doc_sgl/article/details/42344441 考虑n!的质数因子.后缀0总是由质因子2和质因子5相乘得来的.如果我们可以计数2和5的个数,问题就解决了.考虑下面的例子: n = 5…

【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 *…

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 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

172. Factorial Trailing Zeroes Easy 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: You…

LeetCode172 Factorial Trailing Zeroes. LeetCode258 Add Digits. LeetCode268 Missing Number

数学题 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. (Easy) 分析:求n的阶乘中末位0的个数,也就是求n!中因数5的个数(2比5多),简单思路是遍历一遍,对于每个数,以此除以5求其因数5的个数,但会超时. 考虑到一个数n比他小…

[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…

【原创】leetCodeOj --- Factorial Trailing Zeroes 解题报告

原题地址: https://oj.leetcode.com/problems/factorial-trailing-zeroes/ 题目内容: Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 方法: 数学原理很简单,稍微讲一下 我们知道,一堆数相乘出了0,除了有0之外,需要一个2,5数对.比如4…