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Commit bb9b599

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feat: add solutions to lcci problem: No.16.05 (doocs#1474)
No.16.05.Factorial Zeros
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10 files changed

+215
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lines changed

‎lcci/16.05.Factorial Zeros/README.md‎

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## 题目描述
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<!-- 这里写题目描述 -->
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<p>设计一个算法,算出 n 阶乘有多少个尾随零。</p>
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<p><strong>示例 1:</strong></p>
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<pre><strong>输入:</strong> 3
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:数学**
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题目实际上是求 $[1,n]$ 中有多少个 5ドル$ 的因数。
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我们以 130ドル$ 为例来分析:
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1. 第 1ドル$ 次除以 5ドル,ドル得到 26ドル,ドル表示存在 26ドル$ 个包含因数 5ドル$ 的数;
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1. 第 2ドル$ 次除以 5ドル,ドル得到 5ドル,ドル表示存在 5ドル$ 个包含因数 5ドル^2$ 的数;
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1. 第 3ドル$ 次除以 5ドル,ドル得到 1ドル,ドル表示存在 1ドル$ 个包含因数 5ドル^3$ 的数;
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1. 累加得到从 $[1,n]$ 中所有 5ドル$ 的因数的个数。
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时间复杂度 $O(\log n),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def trailingZeroes(self, n: int) -> int:
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ans = 0
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while n:
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n //= 5
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ans += n
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int trailingZeroes(int n) {
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int ans = 0;
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while (n > 0) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int trailingZeroes(int n) {
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int ans = 0;
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while (n) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func trailingZeroes(n int) int {
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ans := 0
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for n > 0 {
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n /= 5
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ans += n
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function trailingZeroes(n: number): number {
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let ans = 0;
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while (n > 0) {
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n = Math.floor(n / 5);
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ans += n;
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}
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return ans;
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}
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```
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### **...**

‎lcci/16.05.Factorial Zeros/README_EN.md‎

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### **Python3**
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```python
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class Solution:
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def trailingZeroes(self, n: int) -> int:
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ans = 0
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while n:
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n //= 5
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ans += n
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int trailingZeroes(int n) {
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int ans = 0;
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while (n > 0) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int trailingZeroes(int n) {
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int ans = 0;
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while (n) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func trailingZeroes(n int) int {
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ans := 0
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for n > 0 {
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n /= 5
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ans += n
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function trailingZeroes(n: number): number {
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let ans = 0;
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while (n > 0) {
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n = Math.floor(n / 5);
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ans += n;
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}
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return ans;
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}
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```
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### **...**
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class Solution {
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public:
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int trailingZeroes(int n) {
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int ans = 0;
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while (n) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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};
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func trailingZeroes(n int) int {
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ans := 0
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for n > 0 {
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n /= 5
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ans += n
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}
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return ans
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}
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class Solution {
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public int trailingZeroes(int n) {
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int ans = 0;
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while (n > 0) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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}
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class Solution:
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def trailingZeroes(self, n: int) -> int:
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ans = 0
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while n:
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n //= 5
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ans += n
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return ans
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function trailingZeroes(n: number): number {
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let ans = 0;
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while (n > 0) {
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n = Math.floor(n / 5);
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ans += n;
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}
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return ans;
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}

‎solution/0100-0199/0172.Factorial Trailing Zeroes/README.md‎

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1. 第 3ドル$ 次除以 5ドル,ドル得到 1ドル,ドル表示存在 1ドル$ 个包含因数 5ドル^3$ 的数;
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1. 累加得到从 $[1,n]$ 中所有 5ドル$ 的因数的个数。
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时间复杂度 $O(\log n),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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}
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```
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### **TypeScript**
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```ts
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function trailingZeroes(n: number): number {
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let ans = 0;
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while (n > 0) {
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n = Math.floor(n / 5);
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ans += n;
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}
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return ans;
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int trailingZeroes(int n) {
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int ans = 0;
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for (int i = 5; i <= n; i *= 5) ans += n / i;
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while (n) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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};
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}
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```
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### **TypeScript**
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```ts
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function trailingZeroes(n: number): number {
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let ans = 0;
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while (n > 0) {
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n = Math.floor(n / 5);
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ans += n;
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}
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return ans;
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}
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```
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### **...**
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```

‎solution/0100-0199/0172.Factorial Trailing Zeroes/README_EN.md‎

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}
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```
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### **TypeScript**
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```ts
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function trailingZeroes(n: number): number {
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let ans = 0;
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while (n > 0) {
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n = Math.floor(n / 5);
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ans += n;
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}
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return ans;
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int trailingZeroes(int n) {
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int ans = 0;
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for (int i = 5; i <= n; i *= 5) ans += n / i;
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while (n) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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};
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}
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```
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### **TypeScript**
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```ts
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function trailingZeroes(n: number): number {
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let ans = 0;
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while (n > 0) {
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n = Math.floor(n / 5);
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ans += n;
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}
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return ans;
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}
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```
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### **...**
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```

‎solution/0100-0199/0172.Factorial Trailing Zeroes/Solution.cpp‎

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public:
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int trailingZeroes(int n) {
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int ans = 0;
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for (int i = 5; i <= n; i *= 5) ans += n / i;
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while (n) {
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n /= 5;
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ans += n;
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}
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return ans;
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}
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};

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