【LeetCode】Factorial Trailing Zeroes

n久不做题了 ,之前因为考研,然后又是假期,一直懒得做,今天开始吧

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

开始没有看到是阶乘,之后又研究复杂度的问题

代码如下:

class Solution {

public:

    int trailingZeroes(int n) {

        int count = 0;

        for (;n > 4;)

        {

            n = n / 5;

            count = count + n;

        }

        return count;

    }

};
时间: 2024-12-21 03:55:59

【LeetCode】Factorial Trailing Zeroes的相关文章

【LeetCode】Factorial Trailing Zeroes (2 solutions)

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*

【leetcode】Factorial Trailing Zeroes(easy)

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) { int ans = 0; while(n > 0) { ans += n / 5; n /= 5; } return ans; }

【Leetcode】Set Matrix Zeroes

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. Could you devise a constant space solution? 思路:因为需要遍历整个矩阵,时间复杂度肯定需要O(m * n),对于空间复杂度而言,第一种是可以使用O(m * n),对每个位置的0的情况进行记录,第二种是使用O(m + n),对每行每列是否存在0进行记录,第三种是O(1)

[LeetCode]172.Factorial Trailing Zeroes

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,不足取. O(logn)解法: 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

LeetCode Day4——Factorial Trailing Zeroes

1 /* 2 * Problem 172: Factorial Trailing Zeroes 3 * Given an integer n, return the number of trailing zeroes in n!. 4 * Note: Your solution should be in logarithmic time complexity. 5 */ 6 7 /* 8 * Solution 1 9 * 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

Java [Leetcode 172]Factorial Trailing Zeroes

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)

Java for LeetCode 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. 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

LeetCode 172 Factorial Trailing Zeroes(阶乘后的零)(*)

翻译 给定一个整型n,返回n!后面的零的个数. 注意:你的解决方案应该在log时间复杂度内. 原文 Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 分析 起初我看题目的时候没太注意,还以为就是求n这个数后面的零而已,虽然心想不会这么简单吧--就写了一份代码提交了,结果WA提示我5的话应该返回1,

leetcode——172 Factorial Trailing Zeroes(N!尾巴上有多少个0,算法复杂度为lg)

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity.(您的解决方案应该在对数时间复杂度.) Hide Tags: Math 题目要求:给定N,求N!的末尾有多少0.要求算法复杂度为lg 解题思路: 思路一: 想的比较简单,先实用for循环进行阶乘运算,然后mod10计算0的个数,但是在OJ检查时,超时