https://oj.leetcode.com/problems/factorial-trailing-zeroes/ 题目内容: Given an integern, return the number of trailing zeroes inn!. Note: Your solution should be in logarithmic time complexity. 方法: 数学原理很简单,稍微讲一下 我们知道,一堆数相乘出了0,除了有0之外,需要一个2,5数对。比如4 ...
LeetCode172——阶乘后的零 /factorial-trailing-zeroes/description/题目描述: 知识点:数学 思路:只有2和5相乘末尾才可能产生0 还有一个事实是,[1, n]范围内2的数量远远大于5的数量,因此有多少个5...; 因此我们的结果就是: 注意,虽然25中包含了2个5,但其中一个5已经在n / 5中被计算过,后续同理。 JAVA...
Question: https://oj.leetcode.com/problems/factorial-trailing-zeroes/ 解答: 这题其实是求小于N的数中是5的倍数的个数。还要注意有些数的因子可能是多个5的情况,比如25,50,所以需要计算25的倍数的个数,同样需要计算125的倍数的个数,所有这些个数的总和就是我们要的结果。 classSolution {public:inttrailing...
Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 思路: 打印结果,然后找规律,发现每隔5个周期0的个数增一,隔5*5再上一轮的基础上再加一,直到5^x>=n。。。 事实上我做的 时候用leetcode的testcase上尝试打印了5、10、...
172. Factorial Trailing Zeroes code class Solution { public: int trailingZeroes(int n) { int ans = 0; while(n) { n /= 5; ans += n; } return ans; } }; 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. ...
Can you solve this real interview question? Factorial Trailing Zeroes - Given an integer n, return the number of trailing zeroes in n!. Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1. Example 1: Input: n = 3 Output: 0 Explanation: 3! = 6,
* 每隔5个数,出现一次5,也就是n / 5 * 每隔25个数,额外出现一次5,也是就是n / 25 * 每隔125个数,额外出现一次5,也就是n / 125 *…… 以此类推,所以代码如下: javapublic class Solution { public int trailingZeroes(int n) { int x = 5; ...
Givenanintegern,returnthenumberoftrailing zeroesinn!. Note: Your solution should beinlogarithmictimecomplexity. 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,这我就纳闷了,5后面毛的0呐……赶紧看题目……...
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2015-07-10 21:04 − 定义: 一个正整数的阶乘(英语:factorial)是所有小于及等于该数的正整数的积,并且有0的阶乘为1。自然数n的阶乘写作n!。 亦即n!=1×2×3×...×n。阶乘亦... WhyWin 0 1484 [LeetCode] Factorial Trailing Zeroes 阶乘末尾0 2015-01-08 22:57 − Given an integer n,...