# Nicksxs's Blog
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## spark-little-tips

Posted on 2017-03-28 Edited on 2020-01-12 In data analysis Views: Disqus:

## spark 的一些粗浅使用经验

工作中学习使用了一下Spark做数据分析,主要是用spark的python接口，首先是`pyspark.SparkContext(appName=xxx)`，这是初始化一个Spark应用实例或者说会话，不能重复，
返回的实例句柄就可以调用`textFile(path)`读取文本文件，这里的文本文件可以是HDFS上的文本文件，也可以普通文本文件，但是需要在Spark的所有集群上都存在，否则会
读取失败，`parallelize`则可以将python生成的集合数据读取后转换成rdd(A Resilient Distributed Dataset (RDD),一种spark下的基本抽象数据集),基于这个RDD就可以做
数据的流式计算，例如`map reduce`,在Spark中可以非常方便地实现

### 简单的mapreduce word count示例

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    textFile = sc.parallelize([(1,1), (2,1), (3,1), (4,1), (5,1),(1,1), (2,1), (3,1), (4,1), (5,1)])
    data = textFile.reduceByKey(lambda x, y: x + y).collect()
    for _ in data:
        print(_)

### 结果

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    (3, 2)
    (1, 2)
    (4, 2)
    (2, 2)
    (5, 2)

## php-abstract-class-and-interface

Posted on 2016-11-10 Edited on 2020-01-12 In php Views: Disqus:

## PHP抽象类和接口

- 抽象类与接口
- 抽象类内可以包含非抽象函数，即可实现函数
- 抽象类内必须包含至少一个抽象方法，抽象类和接口均不能实例化
- 抽象类可以设置访问级别，接口默认都是public
- 类可以实现多个接口但不能继承多个抽象类
- 类必须实现抽象类和接口里的抽象方法，不一定要实现抽象类的非抽象方法
- 接口内不能定义变量，但是可以定义常量

## 示例代码

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    <?php
    interface int1{
        const INTER1 = 111;
        function inter1();
    }
    interface int2{
        const INTER1 = 222;
        function inter2();
    }
    abstract class abst1{
        public function abstr1(){
            echo 1111;
        }
        abstract function abstra1(){
            echo 'ahahahha';
        }
    }
    abstract class abst2{
        public function abstr2(){
            echo 1111;
        }
        abstract function abstra2();
    }
    class normal1 extends abst1{
        protected function abstr2(){
            echo 222;
        }
    }

## result

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    PHP Fatal error:  Abstract function abst1::abstra1() cannot contain body in new.php on line 17

    Fatal error: Abstract function abst1::abstra1() cannot contain body in php on line 17

## summary-ranges-228

Posted on 2016-10-12 Edited on 2020-01-12 In leetcode Views: Disqus:

#### problem

Given a sorted integer array without duplicates, return the summary of its ranges.

For example, given `[0,1,2,4,5,7]`, return `["0->2","4->5","7"]`.

#### 题解

每一个区间的起点`nums[i]`加上`j`是否等于`nums[i+j]`
参考

#### Code

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    class Solution {
    public:
        vector<string> summaryRanges(vector<int>& nums) {
            int i = 0, j = 1, n;
            vector<string> res;
            n = nums.size();
            while(i < n){
                j = 1;
                while(j < n && nums[i+j] - nums[i] == j) j++;
                res.push_back(j <= 1 ? to_string(nums[i]) : to_string(nums[i]) + "->" + to_string(nums[i + j - 1]));
                i += j;
            }
            return res;
        }
    };

## minimum-size-subarray-sum-209

Posted on 2016-10-11 Edited on 2020-01-12 In leetcode Views: Disqus:

### problem

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn’t one, return 0 instead.

For example, given the array `[2,3,1,2,4,3]` and `s = 7`,
the subarray `[4,3]` has the minimal length under the problem constraint.

#### 题解

参考，滑动窗口，跟之前Data Structure课上的online算法有点像，链接

#### Code

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    class Solution {
    public:
        int minSubArrayLen(int s, vector<int>& nums) {
            int len = nums.size();
            if(len == 0) return 0;
            int minlen = INT_MAX;
            int sum = 0;

            int left = 0;
            int right = -1;
            while(right < len)
            {
                while(sum < s && right < len)
                    sum += nums[++right];
                if(sum >= s)
                {
                    minlen = minlen < right - left + 1 ? minlen : right - left + 1;
                    sum -= nums[left++];
                }
            }
            return minlen > len ? 0 : minlen;
        }
    };

## binary-watch

Posted on 2016-09-29 Edited on 2020-01-12 In leetcode Views: Disqus:

### problem

A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59).

Each LED represents a zero or one, with the least significant bit on the right.

For example, the above binary watch reads “3:25”.

Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.

#### Example:

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    Input: n = 1
    Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]

#### Note:

- The order of output does not matter.
- The hour must not contain a leading zero, for example “01:00” is not valid, it should be “1:00”.
- The minute must be consist of two digits and may contain a leading zero, for example “10:2” is not valid, it should be “10:02”.

### 题解

又是参（chao）考（xi）别人的代码，嗯，就是这么不要脸，链接

#### Code

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    class Solution {
    public:
        vector<string> readBinaryWatch(int num) {
            vector<string> res;
            for (int h = 0; h < 12; ++h) {
                for (int m = 0; m < 60; ++m) {
                    if (bitset<10>((h << 6) + m).count() == num) {
                        res.push_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m));
                    }
                }
            }
            return res;
        }
    };