题目 Given an integer n, return the number of trailing zeroes in n!.Note: Your solution should be in logarithmic time complexity. 思路一:不能AC 思路,统计2和5的个数 ,时间复杂度为O(n) int min(int a,int b){return a<b?a:
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,
Problem: Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. Analysis: 对n!做质因数分解n!=2x*3y*5z*… 显然0的个数等于min(x,z),并且min(x,z
Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 出现0的情况是,出现5和2的倍数。 [n/k]代表1~n中能被k整除的个数,而能被2整除的个数多余能被5整除的个数,故只要知
Given an integer n, return the number of trailing zeroes in n!. 题目的意思是要求一个整数的阶乘末尾有多少个0; 1.需要注意的是后缀0是由2,5相乘得来,因此只需看有多少个2,5即可 n = 5: 5!的质因子中 (2 * 2 * 2 * 3 * 5)包含一个5和三个2。因而后缀0的个数是1。 n = 11: 11!的质因子中
Here are examples: a = '12y'# ch = ord(a) TypeError: ord() expected a character, but string of length 3 foundch = chr(ord(a[0])+1)print(ch) #2ch = chr(ord(a[2])+1)print(ch) #z#给字符串赋值没有太好的方式,切片是有效
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 朴素解法: 首先求出n!,然后计算末尾0的个数。(重复÷10,直到余数非0) 该解法在输入的数字稍大时就会导致阶乘得数溢出,
Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 我们对n!分解质因子,n!=2^x+3^y+5^z+...,结尾的0一定是1个2和1个5相乘得到的。很显然,这里因子5的个数小于因子2的个
Given an integer n, return the number of trailing zeroes in n!. Example 1: Input: 3Output: 0Explanation: 3! = 6, no trailing zero. Example 2: Input: 5Output: 1Explanation: 5! = 120, one traili