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更新 0151.反转字符串中的单词 排版格式修复
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@ -28,10 +28,11 @@
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输出: "example good a"
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解释: 如果两个单词间有多余的空格,将反转后单词间的空格减少到只含一个。
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## 算法公开课
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# 思路
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[字符串复杂操作拿捏了! | LeetCode:151.翻转字符串里的单词](https://www.bilibili.com/video/BV1uT41177fX),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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针对本题,我录制了视频讲解:[字符串复杂操作拿捏了! | LeetCode:151.翻转字符串里的单词](https://www.bilibili.com/video/BV1uT41177fX),结合本题解一起看,事半功倍!
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## 思路
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**这道题目可以说是综合考察了字符串的多种操作。**
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@ -204,8 +205,7 @@ public:
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## 其他语言版本
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Java:
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### Java:
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```Java
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class Solution {
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@ -433,9 +433,10 @@ class Solution {
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}
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```
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python:
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### python:
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(版本一)先删除空白,然后整个反转,最后单词反转。
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**因为字符串是不可变类型,所以反转单词的时候,需要将其转换成列表,然后通过join函数再将其转换成列表,所以空间复杂度不是O(1)**
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```Python
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class Solution:
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def reverseWords(self, s: str) -> str:
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@ -467,7 +468,7 @@ class Solution:
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return " ".join(words)
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```
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Go:
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### Go:
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版本一:
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@ -571,7 +572,8 @@ func reverse(b *[]byte, left, right int) {
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javaScript:
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### JavaScript:
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```js
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/**
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* @param {string} s
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@ -630,7 +632,7 @@ function reverse(strArr, start, end) {
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}
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```
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TypeScript:
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### TypeScript:
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```typescript
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function reverseWords(s: string): string {
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@ -689,7 +691,7 @@ function reverseWords(s: string): string {
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};
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```
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Swift:
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### Swift:
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```swift
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func reverseWords(_ s: String) -> String {
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@ -766,7 +768,7 @@ func reverseWord(_ s: inout [Character]) {
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}
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```
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Scala:
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### Scala:
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```scala
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object Solution {
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@ -824,8 +826,8 @@ object Solution {
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}
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```
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### PHP:
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PHP:
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```php
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function reverseWords($s) {
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$this->removeExtraSpaces($s);
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@ -872,7 +874,7 @@ function reverseString(&$s, $start, $end) {
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return ;
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}
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```
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Rust:
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### Rust:
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```Rust
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// 根据C++版本二思路进行实现
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@ -924,7 +926,7 @@ pub fn remove_extra_spaces(s: &mut Vec<char>) {
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}
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}
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```
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C:
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### C:
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```C
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// 翻转字符串中指定范围的字符
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@ -972,3 +974,4 @@ char * reverseWords(char * s){
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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