Normally, the factorial of a positive integer n is the product of all positive integers less than or equal to n. For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. We instead make a clumsy factorial: using the integers in decreasin…
class Solution(object): def clumsy(self, N): """ :type N: int :rtype: int """ op = {0: '*', 1: '//', 2: '+', 3: '-'} strN = list(map(str, range(N - 1, 0, -1))) ret = str(N) for i, s in enumerate(strN): ret += op[i % 4] + s re…
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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,…
Let f(x) be the number of zeroes at the end of x!. (Recall that x! = 1 * 2 * 3 * ... * x, and by convention, 0! = 1.) For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has 2 zeroes at the end. Given…
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…
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 解题方法 直接eval 日期 题目地址:https://leetcode.com/problems/clumsy-factorial/ 题目描述 Normally, the factorial of a positive integer n is the product of all positive integers less than or equal t…
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. 题目标签:Math 题目要求我们找到末尾0的数量. 只有当有10的存在,才会有0,比如 2 * 5 = 10; 4 * 5 = 20; 5 * 6 = 30; 5 * 8 = 40 等等,可以发现0 和 5 的联系. 所以这一题也是在问 n 里有多…
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. 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…
题目如下: Normally, the factorial of a positive integer n is the product of all positive integers less than or equal to n. For example, factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. We instead make a clumsy factorial: using the integers in dec…
