这次比赛之后，下次再遇到O(n²)的题目，而且n < 1000的时候，一定要用C++！这是泪的教训啊😭

第一题Degree of an Array

在统计每个数字频率的同时，记录这个数字出现的最左和最右位置(l, r)。那么包含这个数字的最短数组的长度为r-l+1。

class Solution(object):
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        cnt = {}
        f = 0
        for i, n in enumerate(nums):
            if n not in cnt:
                cnt[n] = [1, i, i]
            else:
                c, a, _ = cnt[n]
                cnt[n] = [c+1, a, i]
            f = max(f, cnt[n][0])
        return min(v[2]-v[1]+1 for v in cnt.values() if v[0]==f)

这次比赛又是对Python不友好。最后一题同样的实现Python会TLE，但是C++就AC。不过也有可能我的算法不是太好。值得一提的是，我发现比赛时我第三题的算法是错误的，但是还是过了。

第一题Binary Number with Alternating Bits

比较直接，代码如下。

class Solution(object):
    def hasAlternatingBits(self, n):
        """
        :type n: int
        :rtype: bool
        """
        b = bin(n)[2:]
        return all(b[i]!=b[i+1] for i in range(len(b)-1))

This article is my notes on support vector machine for Lecture 7 and 8 of Machine Learning by Andrew Ng.

Intuition

In a binary classification problem, we can use logistic regression

where \(g\) is the sigmoid function with a figure of it below.

Then given input \(x\), the model predicts \(1\) if and only if \(\theta^x \ge 0\), in which case \(h_\theta(x) = g(\theta^T x) \ge 0.5\); and it predicts \(0\) if and only if \(\theta^T x < 0\). Moreover, based on the shape of sigmoid function, if \(\theta^T x >> 0\), we are very confident that \(y=1\). Likewise, if \(\theta^T x << 0\), we are very confident that \(y=0\). Therefore, we hope that for the training set \(\{(x^{(i)}, y^{(i)})\}_{i=1}^m\), we can find such a \(\theta\) that \(\theta^T x^{(i)} >> 0\) if \(y^{(i)}=1\) and \(\theta^T x^{(i)} << 0\) if \(y^{(i)}=0\).

This article is my notes on generative model for Lecture 5 and 6 of Machine Learning by Andrew Ng. What we do in logistic regression using generalized linear model is that, we approximate \(P(y|x)\) using given data. This kind of learning algorithms is discriminative, in which we predict \(y\) based on the input features \(x\). On the contrary, generative model is to model \(P(x|y)\), the probability of the features \(x\) given class \(y\). In other words, we want to study how the features structure looks like given a class \(y\). If we also learn what \(P(y)\) is, we can easily recover \(P(y|x)\), for example, in the binary classification problem,

where \(P(x) = P(x|y=0)P(y=0) + P(x|y=1)P(y=1)\).

In this article, we are going to see a simple example of generative model on Gaussian discriminant analysis and Naive Bayes.

这次比赛跪了。。

第一题Repeated String Match

二分查找。首先算出最少需要多少次才有可能使得B是子字符串：必须要保证重复后的A的长度不小于B的长度。我们可以简单算出最少需要重复：

这题竟然WA了2次(因为typo)，TLE一次(一开始用j＝10i)。。。

class Solution(object):
    def repeatedStringMatch(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: int
        """
        i = (len(A) + len(B) - 1) / len(A)
        j = 2 * i
        ans = -1
        while i <= j:
            k = (i+j) / 2
            tmp = A * k
            if tmp.find(B) != -1:
                ans = k
                j = k - 1
            else:
                i = k + 1
        return ans

This article is a companion article to my another post Generalized Linear Model. In this article, I will implement some of the learning algorithms in Generalized Linear Model. To be more specific, I will do some examples on linear regression and logistic regression. With some effort, google search gives me some very good example data sets to work with. The datasets collected by Larry Winner is one of the excellent sets, which will be used in the article.

The implementations here use Python. Required 3rd party libraries are:

• Requests: use to get online data
• Matplotlib: use for plotting
• NumPy: use for matrix calculation

这次比赛做得还算可以，只是最后一题看错题目WA了3次😭

第一题Baseball Game

class Solution(object):
    def calPoints(self, ops):
        """
        :type ops: List[str]
        :rtype: int
        """
        points = []
        for op in ops:
            if op == 'C':
                points.pop()
            elif op == 'D':
                points.append(points[-1]*2)
            elif op == '+':
                points.append(points[-1]+points[-2])
            else:
                points.append(int(op))
        return sum(points)

This article on Generalized Linear Model (GLM) is based on the first four lectures of Machine Learning by Andrew Ng. But the structure of the article is quite different from the lecture. I will talk about exponential family of distributions first. Then I will discuss the general idea of GLM. Finally, I will try to derive some well known learning algorithms from GLM.

Exponential Family

\(\eta\) is called the natural parameter and \(T(y)\) is called the sufficient statistic.

一个网络爬虫大致可以分成三个部分：获取网页，提取信息以及保存信息。Python有很多爬虫框架，其中最出名要数Scrapy了。这也是我唯一用过的Python爬虫框架，用起来很省心。让我苦恼的是，Scrapy在我的Raspberry Pi Zero W安装起来很麻烦，而且我觉得我爬取的网页比较容易处理，没有必要用这么重量级的框架。抱着学习的心态，我开始自己造轮子了。

在造轮子之前我找到些轻量级的框架，Sukhoi是我比较喜欢的一个。该库作者iogf使用了自己的异步库、网络库来写这个框架。这让我很佩服他。我写的框架受到了Sukhoi很大的启发与影响。

这次比赛对python不友好啊，第三题在比赛的时候一直TLE，比赛结束后用C++实现同样的算法就过了😭

第一题Longest Continuous Increasing Subsequence

扫描一遍数组即可：如果nums[i], nums[i+1], ..., nums[j-1]是当前最长严格递增子数组，下次扫描的时候可以直接从j开始。

class Solution(object):
    def findLengthOfLCIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        ans = i = 0
        while i < len(nums):
            j = i + 1
            while j < len(nums) and nums[j] > nums[j-1]:
                j += 1
            ans = max(ans, j-i)
            i = j
        return ans