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Commit cdee514

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feat: add module 'High frequency interview questions'
添加面试高频题模块
1 parent 2d06934 commit cdee514

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9 files changed

+273
-36
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9 files changed

+273
-36
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‎README.md‎

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@@ -31,6 +31,29 @@
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- [LeetCode 《剑指 Offer(第 2 版)》](/lcof/README.md)
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- [LeetCode 《程序员面试金典(第 6 版)》](/lcci/README.md)
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## 面试高频题
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### 数组
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### 链表
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1. [从尾到头打印链表](./lcof/面试题06.%20从尾到头打印链表/README.md)
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1. [删除链表的节点](./lcof/面试题18.%20删除链表的节点/README.md)
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1. [链表中倒数第 k 个节点](./lcof/面试题22.%20链表中倒数第k个节点/README.md)
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1. [反转链表](./solution/0200-0299/0206.Reverse%20Linked%20List/README.md)
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1. [环形链表](./solution/0100-0199/0141.Linked%20List%20Cycle/README.md)
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1. [环形链表 II](./solution/0100-0199/0142.Linked%20List%20Cycle%20II/README.md)
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### 二叉树
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### 数学
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### 栈和队列
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### 动态规划
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### 混合问题
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## 维护者
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[Yang Libin](https://github.com/yanglbme): GitHub 技术社区 [@Doocs](https://github.com/doocs) 创建者;[@TheAlgorithms](https://github.com/TheAlgorithms) 组织成员。

‎README_EN.md‎

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- [LCOF: *Coding Interviews, 2nd Edition*](/lcof/README_EN.md)
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- [LCCI: *Cracking the Coding Interview, 6th Edition*](/lcci/README_EN.md)
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## High frequency interview questions
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### Arrays
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### Linked List
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1.
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### Binary Tree
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### Math
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### Stack & Queue
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### Dynamic Programming
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### Misc
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## Maintainer
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[Yang Libin](https://github.com/yanglbme): Creator of [@Doocs](https://github.com/doocs) technical community; member of [@TheAlgorithms](https://github.com/TheAlgorithms) organization.

‎lcof/面试题24. 反转链表/README.md‎

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### **Go**
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```Go
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```go
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func reverseList(head *ListNode) *ListNode {
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if head == nil ||head.Next == nil {
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return head

‎solution/0200-0299/0206.Reverse Linked List/README.md‎

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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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定义指针 `p`、`q` 分别指向头节点和下一个节点,`pre` 指向头节点的前一个节点。
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遍历链表,改变指针 `p` 指向的节点的指向,将其指向 `pre` 指针指向的节点,即 `p.next = pre`。然后 `pre` 指针指向 `p`,`p`、`q` 指针往前走。
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当遍历结束后,返回 `pre` 指针即可。
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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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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def reverseList(self, head: ListNode) -> ListNode:
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pre, p = None, head
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while p:
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q = p.next
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p.next = pre
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pre = p
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p = q
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return pre
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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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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode reverseList(ListNode head) {
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ListNode pre = null;
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ListNode p = head;
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while (p != null) {
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ListNode q = p.next;
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p.next = pre;
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pre = p;
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p = q;
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}
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return pre;
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}
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}
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```
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### **JavaScript**
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```js
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} head
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* @return {ListNode}
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*/
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var reverseList = function(head) {
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let node = head
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let pre = null
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while(node) {
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let cur = node
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node = cur.next
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cur.next = pre
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pre = cur
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}
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return pre
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};
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```
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### **Go**
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```Go
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func reverseList(head *ListNode) *ListNode {
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if head == nil ||head.Next == nil {
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return head
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}
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dummyHead := &ListNode{}
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cur := head
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for cur != nil {
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tmp := cur.Next
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cur.Next = dummyHead.Next
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dummyHead.Next = cur
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cur = tmp
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}
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return dummyHead.Next
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}
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```
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### **...**
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```
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```

‎solution/0200-0299/0206.Reverse Linked List/README_EN.md‎

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<!-- tabs:start -->
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### **Python3**
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def reverseList(self, head: ListNode) -> ListNode:
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pre, p = None, head
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while p:
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q = p.next
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p.next = pre
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pre = p
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p = q
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return pre
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```
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### **Java**
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```java
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode reverseList(ListNode head) {
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ListNode pre = null;
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ListNode p = head;
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while (p != null) {
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ListNode q = p.next;
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p.next = pre;
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pre = p;
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p = q;
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}
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return pre;
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}
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}
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```
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### **JavaScript**
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```js
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} head
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* @return {ListNode}
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*/
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var reverseList = function(head) {
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let node = head
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let pre = null
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while(node) {
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let cur = node
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node = cur.next
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cur.next = pre
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pre = cur
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}
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return pre
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};
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```
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### **Go**
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```Go
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func reverseList(head *ListNode) *ListNode {
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if head == nil ||head.Next == nil {
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return head
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}
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dummyHead := &ListNode{}
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cur := head
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for cur != nil {
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tmp := cur.Next
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cur.Next = dummyHead.Next
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dummyHead.Next = cur
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cur = tmp
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}
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return dummyHead.Next
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}
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```
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### **...**
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```
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```
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func reverseList(head *ListNode) *ListNode {
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if head == nil ||head.Next == nil {
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return head
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}
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dummyHead := &ListNode{}
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cur := head
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for cur != nil {
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tmp := cur.Next
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cur.Next = dummyHead.Next
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dummyHead.Next = cur
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cur = tmp
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}
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return dummyHead.Next
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}

‎solution/0200-0299/0206.Reverse Linked List/Solution.java‎

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*/
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class Solution {
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public ListNode reverseList(ListNode head) {
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if (head == null) return null;
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ListNode reverse = null;
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while (head != null) {
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ListNode temp = head;
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head = head.next;
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if (reverse == null) {
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reverse = temp;
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temp.next = null;
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} else {
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temp.next = reverse;
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reverse = temp;
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}
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ListNode pre = null;
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ListNode p = head;
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while (p != null) {
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ListNode q = p.next;
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p.next = pre;
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pre = p;
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p = q;
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}
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return reverse;
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return pre;
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}
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}
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const reverseList = function(head){
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if(head == null){
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return head;
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} head
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* @return {ListNode}
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*/
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var reverseList = function(head) {
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let node = head
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let pre = null
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while(node) {
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let cur = node
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node = cur.next
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cur.next = pre
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pre = cur
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}
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let cur = head;
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let pre = null, tmp = null;
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while(cur != null){
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tmp = cur.next;
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cur.next = pre;
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pre = cur;
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cur = tmp;
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}
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return pre;
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}
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return pre
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};
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def reverseList(self, head: ListNode) -> ListNode:
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pre, p = None, head
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while p:
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q = p.next
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p.next = pre
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pre = p
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p = q
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return pre

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