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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 递归 循环 日期 题目描述 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 Expl…

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*…

/* * 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 *…

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…

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…

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…

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]474. Ones and Zeroes 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/ones-and-zeroes/description/ 题目描述: n the computer world, use restricted resource you have to generate maximum benef…

天,这题我已经没有底气高呼“水”了... 题目的地址: https://leetcode.com/problems/sliding-window-maximum/ 题目内容: Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the…

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…

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,…

题目描述: Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 解题思路: 这个题目给的评级是easy,其实只要想到要求n!中0的个数,能够得到0的只有:2,4,5,10,100....而这里面又以5最为稀缺,所以说我们可以得出阶乘的最终结果中的0的数量等于因子中5的数量,比如说10,阶乘含两个0,…

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 解题思路: 计算n能达到的5的最大次幂,算出在这种情况下能提供的5的个数,然后减去之后递归即可,JAVA实现如下: static public int trailingZeroes(int n) { if(n<25) return n/5; lon…

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…

题目描述: Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 解题思路: 对于阶乘而言,也就是1*2*3*...*n[n/k]代表1~n中能被k整除的个数那么很显然[n/2] > [n/5] (左边是逢2增1,右边是逢5增1)[n/2^2] > [n/5^2](左边是逢4增1,右边是逢25增1)…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. (二)解题 题目大意:求n的阶乘算出来的数尾部有多少个0.如5…

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 logarithmic…

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 logarithmic…

题目 Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 分析 Note中提示让用对数的时间复杂度求解,那么如果粗暴的算出N的阶乘然后看末尾0的个数是不可能的. 所以仔细分析,N! = 1 * 2 * 3 * ... * N 而末尾0的个数只与这些乘数中5和2的个数有关,因为每出现一对5和2就会产生…

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 主要是思考清楚计算过程: 将一个数进行因式分解,含有几个5就可以得出几个0(与偶数相乘). 代码很简单. public class Solution { public int trailingZeroes(int n) { int result =…

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 思路:编程之美里有,就是找因子5的个数. int trailingZeroes(int n) { ; ) { ans += n / ; n /= ; } return ans; }…

Given an integer n, return the number of trailing zeroes in n!. 最初的代码 class Solution { public: int trailingZeroes(int n) { long long int fac = 1; int count=0; if (n==0) fac = 1; for(int i = n;i>0;i--) { fac *= i ; } while(fac % 10 == 0) { count ++; f…

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. (1) class Solution { public: int trailingZeroes(int n) { ; ; n / i; i *= ) { ans += n / i; } return ans; } }; (2) class Solu…

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 只有2 * 5才会产生0只要计算出因子中2和5的个数取小的的那个就好了 public class Solution { public int trailingZeroes(int n) { int numsOf2 = 0; int numsOf5…

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 此题是求阶乘后面零的个数. public class Solution { public int trailingZeroes(int n) { int t=0; while(n!=0){ n/=5; t+=n; } return t; } }…

题目: Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 链接: http://leetcode.com/problems/factorial-trailing-zeroes/ 题解: 求n!里有多少个0.其实主要就是看有多少个5,有多少个5就有多少个0,这样我们就可以用一个while循环来搞定.…

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: 题意是要求一个数字的阶乘,末尾有多少个0 要求是对数级别的时间,所以考虑用递归 分析一下,产生一个10,后面加0,找到所有的2*5,或者2的次方×5的次方,任何情况下因子2的个数永远大于5 所以只需要计算因子5的个数,(25*4 =…

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. 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…

QUESTION Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. FIRST TRY class Solution { public: int trailingZeroes(int n) { int divident; ; ; == ) { nOf2++; divident = n/; } ==…