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Algorithm: 每周至少做一个 LeetCode 的算法题
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难度：Easy
题意：Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
示例 1：
Input: [2,2,1]
Output: 1
示例 2：
Input: [4,1,2,1,2]
Output: 4
解法：
本题有多种解法，现给大家讲一种非常容易的思路。
因为本题前提限制了这个数组中每个数字都出现了 2 次，除了那个我们要找的数字。那么我们可以有以下数学方法：
2 * (a + b + c) - (a + a + b + b + c) = c
这样我们就可以找出只出现一次的那个 c 了。但这种方式中要使用的 set() 函数要求使用额外的空间，与 note 中要求的只使用 O(1) 的空间复杂度不符。代码就不写了。
还有一种方法就是使用字典，对每一个元素尝试进行 pop 操作，如果不行，就存入字典中，最后剩下的那个就是我们要找的那个数字了。
代码：
class Solution:
def singleNumber(self, nums: List[int]) -> int:
hash_table = {}
for i in nums:
try:
hash_table.pop(i)
except:
hash_table[i] = 1
return hash_table.popitem()[0]
但这种方式也同样要求使用额外存储空间。
最后一种方法非常简单，但也很难想到，只能用巧妙来形容。本题的题意是找出只出现了一次的那个数字。如果你学过数字电路，那么有一个概念你一定非常熟悉，就是「异或」。一个数字与它本身异或，结果为 0，一个数字与 0 异或，结果为它本身。根据这个思路，答案显而易见。
代码：
class Solution:
def singleNumber(self, nums: List[int]) -> int:
a = 0
for i in nums:
a ^= i
return a
由于并没有额外申请空间，所以空间复杂度为 O(1)，时间复杂度为线性 O(n)。
本文作者是一位老外，他来到中国，感叹于二维码在国内的使用之广泛，于是写了一篇文章列出了 16 个二维码的使用场景。在我看来，也许二维码在我们的日常生活中已经司空见惯，但由一个外国人来介绍这些使用场景，其实能从另一个不同的角度来观察我们的生活，也是一件非常有趣的事。
时间序列分析中，ARMA 模型是非常常用的一种时序建模模型，它理论简单，计算不复杂，拟合效果也很好，所以使用比较广泛。但 ARMA 模型的定阶相对比较麻烦，虽然常用的有 AIC、BIC 等信息准则方式，但手动计算还是略微复杂。为了解决这一问题，1984 年，Tsay 和 Tiao 提出了对 ARMA 定阶的辅助工具 EACF，其结果可以用与 (p, q) 有关的二维表格表示，结果包含由字母「O」组成的三角形，锐角顶点在位置。如下所示：
在 R 中的 TSA 包提供了 eacf(z, ar.max = 7, ma.max = 13) 函数，只需要提供待辨识的时序数据以及最大的阶数范围，即可按照上述表格确定相应阶数。但有一点要注意，本函数默认检验系数非零的检验水平是 0.05，有时不是特别容易看出锐角顶点。
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