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Commitb867080

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206-reverse-linked-list.md Separated Chinese solutions from English solutions.
1 parent0ea30cc commitb867080

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‎en/1-1000/206-reverse-linked-list.md

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#206. Reverse Linked List - Best Practices of LeetCode Solutions
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LeetCode link:[206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list),
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[206. 反转链表](https://leetcode.cn/problems/reverse-linked-list)
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LeetCode link:[206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list), difficulty:**Easy**.
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[中文题解](#中文题解)
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##LeetCode problem description
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##LeetCode description of "206. Reverse Linked List"
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Given the`head` of a singly linked list, reverse the list, and return_the reversed list_.
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###[Example 1]
@@ -31,13 +28,11 @@ Given the `head` of a singly linked list, reverse the list, and return _the reve
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-`-5000 <= Node.val <= 5000`
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##Intuition
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[中文题解](#中文题解)
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1. To solve this problem, we only need to define**two** variables:`current` and`previous`.
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2.`current.next = previous` is the inversion.
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3. The loop condition should be`while (current != null)` instead of`while (current.next != null)`, because the operation to be performed is`current.next = previous`.
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##Steps to the Solution
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##Steps
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1. Traverse all nodes.
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```javascript
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previous=null
@@ -284,55 +279,3 @@ end
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##问题描述
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###[Example 1]
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给你单链表的头节点`head` ，请你反转链表，并返回反转后的链表。
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**输入**:`head = [1,2,3,4,5]`
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**输出**:`[5,4,3,2,1]`
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##中文题解
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###思路
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1. 解决这个问题，只需要定义**两**个变量：`current`和`previous`。
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2.`current.next = previous`就是反转了。
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3. 循环条件应是`while (current != null)`，而不应该是`while (current.next != null)`，因为需要操作的是`current.next = previous`.
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###步骤
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1. 遍历所有节点。
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```javascript
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previous=null
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current= head
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while (current!=null) {
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current=current.next
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}
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```
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2. 加入`current.next = previous`。
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```javascript
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previous=null
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current= head
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while (current!=null) {
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tempNext=current.next
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current.next= previous
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current= tempNext
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}
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```
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3.`previous`目前始终是`null`，需要让它变化起来：`previous = current`。
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```javascript
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previous=null
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current= head
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while (current!=null) {
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tempNext=current.next
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current.next= previous
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previous= current
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current= tempNext
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}
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```

‎en/3001-4000/unorganized.md

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@@ -205,4 +205,4 @@ Add a table to show the differences between A-Start and breadth-first search
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-https://leetcode.com/problems/k-closest-points-to-origin/
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-https://leetcode.com/problems/find-special-substring-of-length-k/
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-https://leetcode.cn/problems/eat-pizzas/
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-
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-https://leetcode.cn/problems/merge-intervals/

‎zh/1-1000/206-reverse-linked-list.md

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#206. Reverse Linked List - Best Practices of LeetCode Solutions
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LeetCode link:[206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list),
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[206. 反转链表](https://leetcode.cn/problems/reverse-linked-list)
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#206. 反转链表 - 力扣题解最佳实践
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力扣链接：[206. 反转链表](https://leetcode.cn/problems/reverse-linked-list) ，难度：**简单**。
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[中文题解](#中文题解)
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##LeetCode problem description
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Given the`head` of a singly linked list, reverse the list, and return_the reversed list_.
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##力扣“206. 反转链表”问题描述
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给你单链表的头节点`head` ，请你反转链表，并返回反转后的链表。
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###[Example 1]
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###[示例 1]
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![](../../images/examples/206_1.jpg)
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**Input**:`head = [1,2,3,4,5]`
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**输入**:`head = [1,2,3,4,5]`
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**Output**:`[5,4,3,2,1]`
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**输出**:`[5,4,3,2,1]`
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###[Example 2]
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###[示例 2]
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![](../../images/examples/206_2.jpg)
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**Input**:`[1,2]`
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**输入**:`[1,2]`
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**Output**:`[2,1]`
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**输出**:`[2,1]`
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###[Example 3]
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**Input**:`[]`
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###[示例 3]
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**输入**:`[]`
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**Output**:`[]`
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**输出**:`[]`
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###[Constraints]
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-The number of nodes in the list is the range`[0, 5000]`.
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###[约束]
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-链表中节点的数目范围是`[0, 5000]`
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-`-5000 <= Node.val <= 5000`
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##Intuition
34-
[中文题解](#中文题解)
35-
36-
1. To solve this problem, we only need to define**two** variables:`current` and`previous`.
37-
2.`current.next = previous` is the inversion.
38-
3. The loop condition should be`while (current != null)` instead of`while (current.next != null)`, because the operation to be performed is`current.next = previous`.
30+
##思路
31+
1. 解决这个问题，只需要定义**两**个变量：`current`和`previous`。
32+
2.`current.next = previous`就是反转了。
33+
3. 循环条件应是`while (current != null)`，而不应该是`while (current.next != null)`，因为需要操作的是`current.next = previous`.
3934

40-
##Steps to the Solution
41-
1.Traverse all nodes.
35+
##步骤
36+
1.遍历所有节点。
4237
```javascript
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previous=null
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current= head
@@ -48,7 +43,7 @@ while (current != null) {
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}
4944
```
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51-
2.Add`current.next = previous`.
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2.加入`current.next = previous`。
5247
```javascript
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previous=null
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current= head
@@ -60,7 +55,7 @@ while (current != null) {
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}
6156
```
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63-
3.`previous` is always`null`, we need to change it:`previous = current`.
58+
3.`previous`目前始终是`null`，需要让它变化起来：`previous = current`。
6459
```javascript
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previous=null
6661
current= head
@@ -73,9 +68,12 @@ while (current != null) {
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}
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```
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##Complexity
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* Time:`O(n)`.
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* Space:`O(1)`.
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##复杂度
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* 时间：`O(N)`。
73+
* 空间：`O(1)`。
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##进阶
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链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？
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##Java
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```java
@@ -260,79 +258,7 @@ def reverse_list(head)
260258
end
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```
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##C
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```c
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Kotlin
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```kotlin
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// Welcome to create a PR to complete the code of this language, thanks!
271-
```
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##Swift
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```swift
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// Welcome to create a PR to complete the code of this language, thanks!
261+
##C, Kotlin, Swift, Rust or other languages
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```
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##Rust
279-
```rust
280263
// Welcome to create a PR to complete the code of this language, thanks!
281264
```
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##Other languages
284-
```
285-
// Welcome to create a PR to complete the code of this language, thanks!
286-
```
287-
288-
##问题描述
289-
290-
291-
###[Example 1]
292-
给你单链表的头节点`head` ，请你反转链表，并返回反转后的链表。
293-
294-
**输入**:`head = [1,2,3,4,5]`
295-
296-
**输出**:`[5,4,3,2,1]`
297-
298-
##中文题解
299-
###思路
300-
1. 解决这个问题，只需要定义**两**个变量：`current`和`previous`。
301-
2.`current.next = previous`就是反转了。
302-
3. 循环条件应是`while (current != null)`，而不应该是`while (current.next != null)`，因为需要操作的是`current.next = previous`.
303-
304-
###步骤
305-
1. 遍历所有节点。
306-
```javascript
307-
previous=null
308-
current= head
309-
310-
while (current!=null) {
311-
current=current.next
312-
}
313-
```
314-
315-
2. 加入`current.next = previous`。
316-
```javascript
317-
previous=null
318-
current= head
319-
320-
while (current!=null) {
321-
tempNext=current.next
322-
current.next= previous
323-
current= tempNext
324-
}
325-
```
326-
327-
3.`previous`目前始终是`null`，需要让它变化起来：`previous = current`。
328-
```javascript
329-
previous=null
330-
current= head
331-
332-
while (current!=null) {
333-
tempNext=current.next
334-
current.next= previous
335-
previous= current
336-
current= tempNext
337-
}
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```

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