从这个类的名字Unsafe上来说这个类就是一个不安全的类，也是不开放给用户直接使用的（当然我们还是可以通过其他一些方法用到）。
这个类在jdk源码中多个类中用到，主要作用是任意内存地址位置处读写数据，外加一下CAS操作。它的大部分操作都是绕过JVM通过JNI完成的，因此它所分配的内存需要手动free，所以是非常危险的。但是Unsafe中很多（但不是所有）方法都很有用，且有些情况下，除了使用JNI，没有其他方法弄够完成同样的事情。
至于研究它的起因，是因为我最近在看jdk8的ConcurrentHashMap，这个版本的主要函数就是用过Unsafe来完成的。
大家都知道Java中String是不可变类。也就是说如果有一个String初始化的值位”Hello”，如果把它赋值为”Hello World!”，那么不是在原内存地址上修改数据，而是重新生成了一个新对象并重新指向这个对象。而在Java虚拟机中，字符串是直接放在常量池中的，而且关于String比较相关的题目很多，这篇文章不做详细描述。
下面的代码会生成几个String，先不要着急的往下看答案，自己想想。1String str = "Hello" + "World" + "!";
这个问题主要还是考察了Java虚拟机对String中”+”运算符的特殊处理，我们来看一下编译生成的class是什么样的。见下图。
看到String HelloWorld!是不是很惊讶，是的，这段代码只生成了一个字符串。
之前一直在用多说评论，但是多说就要关闭了，就是在2017年6月1日。所以必须在这之前改用新的评论系统，不然的话博客就要不能评论了（然而并没有什么人给我评论）。然后就是选择用什么评论系统了，也没有什么可参考，随便选了一个。
至于为什么要自己添加，因为我的Next版本很旧了，但是自己做过很多改动不想更新到最新版本。
In the video game Fallout 4, the quest “Road to Freedom” requires players to reach a metal dial called the “Freedom Trail Ring”, and use the dial to spell a specific keyword in order to open the door.
Given a string ring, which represents the code engraved on the outer ring and another string key, which represents the keyword needs to be spelled. You need to find the minimum number of steps in order to spell all the characters in the keyword.
Initially, the first character of the ring is aligned at 12:00 direction. You need to spell all the characters in the string key one by one by rotating the ring clockwise or anticlockwise to make each character of the string key aligned at 12:00 direction and then by pressing the center button.
At the stage of rotating the ring to spell the key character key[i]:
You can rotate the ring clockwise or anticlockwise one place, which counts as 1 step. The final purpose of the rotation is to align one of the string ring’s characters at the 12:00 direction, where this character must equal to the character key[i].
If the character key[i] has been aligned at the 12:00 direction, you need to press the center button to spell, which also counts as 1 step. After the pressing, you could begin to spell the next character in the key (next stage), otherwise, you’ve finished all the spelling.
Example:
Input: ring = “godding”, key = “gd”
Output: 4
Explanation:
For the first key character ‘g’, since it is already in place, we just need 1 step to spell this character.
For the second key character ‘d’, we need to rotate the ring “godding” anticlockwise by two steps to make it become >”ddinggo”.
Also, we need 1 more step for spelling.
So the final output is 4.
Note:
题目来自于辐射4这个游戏。有一个圆盘，圆盘上有几个英文字母，圆盘12点方向代表当前选中的字母，通过这个圆盘拼出一个和题目一样的字符串，求出最小的步数。
圆盘可以顺时针或者逆时针旋转，每旋转一个字母代表1步，当12点方向是需要的英文字母时，按确定按钮，这个按也算一步。
Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.
Example 1:
Input:
s = “abpcplea”, d = [“ale”,”apple”,”monkey”,”plea”]Output:
“apple”
Example 2:
Input:
s = “abpcplea”, d = [“a”,”b”,”c”]Output:
“a”
Note:
给出一个String A和一个String数组，在String数组中找到最长的String B，这个String可以满足从A中删除一些字母得到。如果满足的字符串有多个，则选择字母序最小的字符串。
Given a list of 24-hour clock time points in “Hour:Minutes” format, find the minimum minutes difference between any two time points in the list.
Example 1:
Input: [“23:59”,”00:00”]
Output: 1
Note:
- The number of time points in the given list is at least 2 and won’t exceed 20000.
- The input time is legal and ranges from 00:00 to 23:59.
计算最短的时间间隔，注意24小时制值得变化。
Note: This is a companion problem to the System Design problem: Design TinyURL.
TinyURL is a URL shortening service where you enter a URL such as https://leetcode.com/problems/design-tinyurl and it returns a short URL such as http://tinyurl.com/4e9iAk.
Design the encode and decode methods for the TinyURL service. There is no restriction on how your encode/decode algorithm should work. You just need to ensure that a URL can be encoded to a tiny URL and the tiny URL can be decoded to the original URL.
完成短链接的编解码。
Let’s play the minesweeper game (Wikipedia), online game)!
You are given a 2D char matrix representing the game board. ‘M’ represents an unrevealed mine, ‘E’ represents an unrevealed empty square, ‘B’ represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit (‘1’ to ‘8’) represents how many mines are adjacent to this revealed square, and finally ‘X’ represents a revealed mine.
Now given the next click position (row and column indices) among all the unrevealed squares (‘M’ or ‘E’), return the board after revealing this position according to the following rules:
If a mine (‘M’) is revealed, then the game is over - change it to ‘X’.
If an empty square (‘E’) with no adjacent mines is revealed, then change it to revealed blank (‘B’) and all of its adjacent unrevealed squares should be revealed recursively.
If an empty square (‘E’) with at least one adjacent mine is revealed, then change it to a digit (‘1’ to ‘8’) representing the number of adjacent mines.
Return the board when no more squares will be revealed.
Example 1:
Input:
[[‘E’, ‘E’, ‘E’, ‘E’, ‘E’],
[‘E’, ‘E’, ‘M’, ‘E’, ‘E’],
[‘E’, ‘E’, ‘E’, ‘E’, ‘E’],
[‘E’, ‘E’, ‘E’, ‘E’, ‘E’]]Click : [3,0]
Output:
[[‘B’, ‘1’, ‘E’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘M’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘1’, ‘1’, ‘B’],
[‘B’, ‘B’, ‘B’, ‘B’, ‘B’]]Explanation:
Example 2:
Input:
[[‘B’, ‘1’, ‘E’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘M’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘1’, ‘1’, ‘B’],
[‘B’, ‘B’, ‘B’, ‘B’, ‘B’]]Click : [1,2]
Output:
[[‘B’, ‘1’, ‘E’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘X’, ‘1’, ‘B’],
[‘B’, ‘1’, ‘1’, ‘1’, ‘B’],
[‘B’, ‘B’, ‘B’, ‘B’, ‘B’]]Explanation:
Note:
这个题指的是扫雷游戏。在扫雷游戏中，用二维矩阵标识当前的扫雷状态。没有被发掘的空块标记为'E'，没有被发掘的雷为'M'，发掘之后的空块标记为'N'，发掘之后的块标记为'X'，另外数字1-8表示该块周围有几个雷。
给出一个二维矩阵标识当天扫雷游戏的状态，然后给出下一步，求出走完下一步之后的游戏状态。
Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 ≤ i ≤ N) in this array:
Example 1:
Input: 2
Output: 2
Explanation:The first beautiful arrangement is [1, 2]:
Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).
Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).
The second beautiful arrangement is [2, 1]:
Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).
Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.
Note:
所谓美丽的排布是指，在i位置上的数字能够整除i或者被i整除，i从1开始。给定一个数字N，计算有多少个这样的排布。
Given a string s, find the longest palindromic subsequence’s length in s. You may assume that the maximum length of s is 1000.
Example 1:
Input:
“bbbab”
Output:
4
One possible longest palindromic subsequence is “bbbb”.
Example 2:
Input:
“cbbd”
Output:
2
One possible longest palindromic subsequence is “bb”.
查找一个字符串中包含的最长回文长度。
