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…

/* * 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 *…

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!. 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…

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. 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!. 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…

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)…

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 =…