程序员子悠 · 好记性不如烂笔头

Introduction

• Algorithm - Learning Algorithm
• Review - Learning English
• Tip - Learning Techniques
• Share - Learning Influence

Let's do it!!!

Algorithm

1. Description

• There are two sorted arrays nums1 and nums2 of size m and n respectively.

• Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

  Example 1:
  nums1 = [1, 3]
  nums2 = [2]

  The median is 2.0
  Example 2:
  nums1 = [1, 2]
  nums2 = [3, 4]

  The median is (2 + 3)/2 = 2.5

2. Solution

package com.silence.arts.leetcode.first;

import java.util.Arrays;

/**
 * <br>
 * <b>Function：</b><br>
 * <b>Author：</b>@author Silence<br>
 * <b>Date：</b>2018-07-02 22:23<br>
 * <b>Desc：</b>无<br>
 */
public class Median_4 {

    /**
     * T System.arraycopy(nums1, 0, sum, 0, nums1.length);
     * @param nums1
     * @param nums2
     * @return
     */
    public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
        double median;
        int[] sum = new int[nums1.length + nums2.length];
        System.arraycopy(nums1, 0, sum, 0, nums1.length);
        System.arraycopy(nums2, 0, sum, nums1.length, nums2.length);

        Arrays.sort(sum);
        //偶数
        if (sum.length % 2 == 0) {
            int a = sum.length / 2 - 1;
            int b = a + 1;
            median = (sum[a] + sum[b]) / 2.0;
            //奇数
        } else {
            median = sum[sum.length / 2.0];
        }
        return median;
    }


    public static void main(String[] args) {
//        int[] a = new int[]{1,9,9,7};
        int[] a = new int[]{};
        int[] b = new int[]{2, 3};
//        int[] b = new int[]{5,3,4,9};
        System.out.println(findMedianSortedArrays(a, b));
    }
}

Thinking

1. combine the two arrays as a new array
2. sort the new array
3. according to the length of the new array decides how to calculate the median
4. if the even length the median is (sum[a] + sum[b]) / 2.0
5. if the odd length the median is sum[array.length / 2.0]

思路

1. 将输入的两个整型数组组合成一个新的数组
2. 将新的数组进行排序
3. 根据新数组的长度进行中位数的计算
4. 如果新数组的长度是偶数，则中位数计算方式为(sum[a] + sum[b]) / 2.0
5. 如果新数组的长度是奇数，则中位数计算方式为sum[array.lenth / 2.0]

Review

What motivates more: positive or negative feedback?

This week I am going to talk about the article What motivates more: positive or negative feedback?

• Motivation is hard to summon and harder to keep alive.

  Often, a sustained blast of it is all we need to kickstart a new career or complete a personal goal. And if we could harness it every day — well, we’d be unstoppable.
  Unfortunately, it’s rarely that simple. Motivation is hard to summon and harder to keep alive. As a resource, it’s as finite as they come (see also: its ever-flakey brother, inspiration).

• It's also positive where there is a negative feedback.

  Negative feedback also can be positive although it's always feeling painful.
 The Positive Power of Negative Thinking

• Regular, scheduled feedback motivates more

  If skillful feedback is about paying attention, then it follows that giving it in small doses, focused on short time spans, is going to be more effective.

Tips

1. Linux统计日志文件的行数

• sed -n '$=' test.log

2. 过滤指定列字段到新文件

• cat test.log |awk -F '&' '{print $7}' > mac.log
• awk -F 指定分隔符，$7 打印出指定列

3. 排序合并，按统计数据排序

• sort mac.log |uniq -c |sort -rn

4. vim和shell切换

• 在 Vi/Vim的命令模式下输入 :sh即可进入 Linux/Unix shell 环境。在要返回到 Vi/Vim 编辑环境时，输入 exit 命令即可。

Share

前段时间项目中使用Consul作为服务器注册和发现的使用，结合Consul-template组件实现服务上下线后，nginx配置文件自动更新的功能。
具体实现见下文。
Consul-template，nginx实现负载均衡和服务列表自动更新