题目: 尾部的零 设计一个算法,计算出n阶乘中尾部零的个数 样例 11! = 39916800,因此应该返回 2 挑战 O(logN)的时间复杂度 解题: 常用方法: 也许你在编程之美中看到,通过求能够被2 整除和能够被5整除个数的最小值就是答案,或者直接求能够被5整除的个数就是答案<能够被5整除的数显然比较小>,但是在这里,java python都试了,结果都会出现运行超时或者越界的问题. 维基百科中有如下计算方法: Java程序: class Solution { /* * param n…

尾部的零:设计一个算法,计算出n阶乘中尾部零的个数 样例:11! = 39916800.因此应该返回2 分析:假如你把1 × 2 ×3× 4 ×……×N中每一个因数分解质因数,例如 1 × 2 × 3 × (2 × 2) × 5 × (2 × 3) × 7 × (2 × 2 ×2) ×…… 10进制数结尾的每一个0都表示有一个因数10存在. 10可以分解为2 × 5,因此只有质数2和5相乘能产生0,别的任何两个质数相乘都不能产生0,而且2,5相乘只产生一个0. 所以,分解后的整个因数式中有多少对…

Write an algorithm which computes the number of trailing zeros in n factorial. Have you met this question in a real interview? Yes Example 11! = 39916800, so the out should be 2 Challenge O(log N) time LeetCode上的原题,请参见我之前的博客Factorial Trailing Zeroes.…

2. Trailing Zeros[easy] Write an algorithm which computes the number of trailing zeros in n factorial. Have you met this question in a real interview? Yes Example 11! = 39916800, so the out should be 2 Challenge O(log N) time 解法一: class Solution { /*…

问题描述: Write a program that will calculate the number of trailing zeros in a factorial of a given number. N! = 1 * 2 * 3 * ... * N Be careful 1000! has 2568 digits... For more info, see: http://mathworld.wolfram.com/Factorial.html Examples zeros(6) =…

Write an algorithm which computes the number of trailing zeros in n factorial. Have you met this question in a real interview? Yes Example 11! = 39916800, so the out should be 2 /* * param n: As desciption * return: An integer, denote the number of t…

Well, to compute the number of trailing zeros, we need to first think clear about what will generate a trailing 0? Obviously, a number multiplied by 10 will have a trailing 0 added to it. So we only need to find out how many 10's will appear in the e…

Description Write an algorithm which computes the number of trailing zeros in n factorial. Example 11! = 39916800, so the out should be 2 Challenge O(log N) time Answer /* * @param n: A long integer * @return: An integer, denote the number of trailin…

计算阶乘尾部的0的个数,初一看很简单. 先上代码 public static long GetFactorial(long n) { || n == ) ; ); } //Main方法中调用 ); ; ; while (true) { ) { num = num * ; count++; } else { break; } } //count就是最终的结果 提交以后才发现问题,计算阶乘的数太大,会导致溢出.查了会资料,用数组存储数字,就不会有溢出的问题了.比如数字120, 存在数组里的结果是 a…

LeetCode上的原题,讲解请参见我之前的博客Factorial Trailing Zeroes. 解法一: int trailing_zeros(int n) { ; while (n) { res += n / ; n /= ; } return res; } 解法二: int trailing_zeros(int n) { ? : n / + trailing_zeros(n / ); } CareerCup All in One 题目汇总…