Netflix Blog - Tips for High Availability Spinnaker介绍
这篇文章Tips for High Availability 是出自 Netflix Technology Blog
这篇文章Tips for High Availability 是出自 Netflix Technology Blog
strace跟踪应用程序时,有标准输出和错误输出,需要将程序执行的输出标准输出和错误输出重定向到一个日志文件中。
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进程出现状态进程,(zombie process )。对于创建多进程的场景, 子进程状态改变产生此信号, 对SIGCLD信号处理不当,会导致僵尸进程。
vmware安装的虚拟机, 今天登录忽然提示“You must change your password now and login again!”, 密码过期了, 我登录设置了ssh免密登录, 输入原始密码,输入新密码后解决。修改成功登录后看了下,密码有效期60天。 通过chage 修改应用用户密码永不过期。
查找initramfs资料的时候看到了这篇文章 ramfs-rootfs-initramfs.txt,
Given a singly linked list, determine if it is a palindrome. Example 1: 1 2 Input: 1->2 Output: false Example 2: 1 2 Input: 1->2->2->1 Output: true Follow up: Could you do it in O(n) time and O(1) space? 解题: 找到中间节点后反转判断节点回文 实现: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 /** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ bool isPalindrome(struct ListNode* head){ if (head == NULL || head->next == NULL) { return true; } struct ListNode* slow=head; struct ListNode* fast=head; struct ListNode* pre=NULL; struct ListNode* next=NULL; //get middle while( fast!=NULL && fast->next!=NULL){ slow = slow->next; fast = fast->next->next; } if(fast) { slow=slow->next; } struct ListNode* cur=slow; //and reverse while(cur!=NULL){ pre = cur; cur = cur->next; pre->next = next; next = pre; } //compare cur = head; while(pre!=NULL){ //printf("%d %d\n", head->val, pre->val); if (cur->val != pre->val){ return false; } cur=cur->next; pre= pre->next; } return true; }
鸟哥解析 CentOS 7.x 的initramfs 档案内容 文章内容时, 主要疑问是为什么使用的是512bytes去算 “initramfs 档案的前置字元容量 ”的大小。
最近在学习 鸟哥私房菜第19章,文章在分析Boot Loader 过程中, 提到到了虚拟档案系统Initial RAM Disk或Initial RAM Filesystem,
题目: Given an array of integers nums, write a method that returns the “pivot” index of this array. We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index. If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index. ...