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2012年，由于项目的需要我第一次接触到了Netty，当时Netty还处于3.x版本。我用十几篇博文记录了自己自学Netty的过程，虽然内容浅薄，但没想到被各处转载，我想主要是因为当时Netty的资料确实较少的缘故。

五六年过去了，Netty早已发展到了4.x系列，好奇也好，求知也罢，我打算重学Netty，虽然严格来说，我已不是IT从业人员，但我仍希望保留对技术的热爱与追求。

学习Netty，就免不了先去了解Java中的几种通信模型。我不想先去学习很多概念，单刀直入，就先从最容易理解的BIO（阻塞I/O）学起。

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Problem

Implement the following operations of a queue using stacks.

push(x) -- Push element x to the back of queue.
pop() -- Removes the element from in front of queue.
peek() -- Get the front element.
empty() -- Return whether the queue is empty.

Example:

MyQueue queue = new MyQueue();

queue.push(1);
queue.push(2);  
queue.peek();  // returns 1
queue.pop();   // returns 1
queue.empty(); // returns false

Notes:

You must use only standard operations of a stack – which means only push to top, peek/pop from top, size, and is empty operations are valid. Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack. You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).false

即用stack来实现queue

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Problem

Given an integer, write a function to determine if it is a power of two.

Example 1:

Input: 1
Output: true 
Explanation: 2^0 = 1

Example 2:

Input: 16
Output: true
Explanation: 2^4 = 16

Example 3:

Input: 218
Output: false

判断一个整数是否是2的幂次方

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Problem

Implement the following operations of a stack using queues.

push(x) – Push element x onto stack. pop() – Removes the element on top of the stack. top() – Get the top element. empty() – Return whether the stack is empty. Example:

MyStack stack = new MyStack();

stack.push(1);
stack.push(2);  
stack.top();   // returns 2
stack.pop();   // returns 2
stack.empty(); // returns false

Notes:

You must use only standard operations of a queue – which means only push to back, peek/pop from front, size, and is empty operations are valid. Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue. You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

利用队列实现栈，这个题真是挺奇怪的。。。所以好像评价不高。。

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Problem

Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

Example 1:

Input: nums = [1,2,3,1], k = 3
Output: true

Example 2:

Input: nums = [1,0,1,1], k = 1
Output: true

Example 3:

Input: nums = [1,2,3,1,2,3], k = 2
Output: false

给定一个整数数组和一个整数 k，判断数组中是否存在两个不同的索引 i 和 j，使得 nums [i] = nums [j]，并且 i 和 j 的差的绝对值最大为 k。

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Problem

Given an array of integers, find if the array contains any duplicates.

Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.

Example 1:

Input: [1,2,3,1]
Output: true

Example 2:

Input: [1,2,3,4]
Output: false

Example 3:

Input: [1,1,1,3,3,4,3,2,4,2]
Output: true

即判断数据中是否有重复元素

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Problem

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

即反转链表，建议采用递归和迭代两种方式。

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Problem

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

Example 1:

Input: s = "egg", t = "add"
Output: true

Example 2:

Input: s = "foo", t = "bar"
Output: false

Example 3:

Input: s = "paper", t = "title"
Output: true

Note: You may assume both s and t have the same length.

同构字符串：

给定两个字符串 s 和 t，判断它们是否是同构的。

如果 s 中的字符可以被替换得到 t ，那么这两个字符串是同构的。

所有出现的字符都必须用另一个字符替换，同时保留字符的顺序。两个字符不能映射到同一个字符上，但字符可以映射自己本身。

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