第一个面试工程师
Dana (Software Engineer) 8:30 - 9:45
很资深的工程师，做过开发和管理， 用过c++， python 之类的开发语言。 现在开始转作java开发， 一开始给我讲了NimbleRx 公司的系统架构和她负责那一块东西。
-
Two sum 变体，用hashMap来做，map.containsKey(sum - arr[i])
Follow up: 如果不能用额外的空间
我说 Arrays.sort, 然后用binary search， 她说，在binary search之前，你怎么sort array。 我就随便说了insert search 。她否定了，还是得用额外的空间，那我又说merge sort ，被否定后。 她说可以用 quick sort -
common father of two nodes in Binary tree
这道题，我之前做过，临时想不起来方法， 只记得要保存访问过的node
在问她要提示以后，勉强做出来了。
第二个面试工程师
Amy (Software Engineer) 9:45 - 1100 (7年工作经验） qa出身，IBM干过， 熟悉db2
主要问了很多测试的问题
例如，Textbox
各种测试的项目
1.test scenario : pharmacy, doctor, patient, driver , 的流程， 如何测试， 怎样保证正确性
take picture 来验证是否收到货
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test an user registration page: username and password, username is an email address, password have a list of limitation
- least 6 characters,
- at least one Capital character,
- 1 number
- 1 low case
- 1 special character
-
test text box: (Country, unicode , utf-8, multi-line,
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test number: ( 360-553-2098, phone number? ) need to be calculate, the range of number ,
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how to evaluate the relation between developer and QA?
5.matrix (code, test cases, test plans)
1).valid test cases
2). invalid test cases
* a. data related vs non data related
* b. severity
* c. UI related or backend API
* d. module, domain,
中途进来一个人 Tyler
不是做技术的。 和他聊了会儿天。带我去取吃的食物。
Lunch (Engineering team) Yulei
lulei 是华人工程师，给我讲了公司的一些情况，一起吃饭，了解了一些相互的情况。 他给我介绍了公司的一些项目和开发的情况: 既做Node.js＋ Express + MongoDB, 又做 Java ＋ spring ＋ Hibernate ＋ mysql， 服务器部署到 AWS 的开发（s3 + Elastic Beanstalk)
第三个面试工程师
Alan (Software Engineer)1:00 - 2:00
出了两道题
- isValidXml (List<Tag> list)
class Tag {
String getTageName(); // return the tag, ex, html
boolean isOpen();
boolean isClose();
}
isValidXml (List<Tag> list ) {
Stack<Tag> st = new Stack<Tag>();
for (Tag tag : list ) {
if ( tag.isOpen()) {
st.push(tag.getTagName());
}
if (tag.isClose() && st.peek() == tag.getTagName()) {
st.pop();
} else {
return false;
}
}
return st.size() == 0;
}
- Connect nodes at same level
问他要了提示， 磕磕巴巴写完程序
// Java program to connect nodes at same level
// NimbleRx coding question
// Date: 2016-05-12
// A binary tree node
class Node {
int val;
Node left, right, next;
Node(int item) {
val = item;
left = null;
right = null;
next = null;
}
}
public class ConnectedTree {
static Node root;
// set the next of root and calls helper recursively for other nodes
void connectNodes(Node p) {
p.next = null; // Set the next for root
// recursively set the next for rest of the nodes
helper(p);
}
// set next of all descendants of p.
public void helper(Node p) {
if (p == null) {
return;
}
if (p.left != null) {
p.left.next = p.right;
}
// set the next node for p's right child
if (p.right != null) {
if (p.next != null) {
p.right.next = p.next.left;
} else {
p.right.next = null; // if p is the right most child at its
// level
}
}
// Set next for other nodes
helper(p.left);
helper(p.right);
}
public static void main(String args[]) {
ConnectedTree tree = new ConnectedTree();
tree.root = new Node(10);
tree.root.left = new Node(8);
tree.root.right = new Node(2);
tree.root.left.left = new Node(1);
tree.root.left.right = new Node(4);
tree.root.right.left = new Node(7);
tree.root.right.right = new Node(3);
// set next nodes in all nodes
tree.connectNodes(root);
// Let us check the values of next nodes
System.out.println("Check the next node in the ConnecteTree " );
System.out.println("Print -1 if there is no next node :\n");
int rootNext = root.next != null ? root.next.val : -1;
System.out.println("next of " + root.val + " is " + rootNext);
int leftNext = root.left.next != null ? root.left.next.val : -1;
System.out.println("next of " + root.left.val + " is " + leftNext);
int rightNext = root.right.next != null ? root.right.next.val : -1;
System.out.println("next of " + root.right.val + " is " + rightNext);
int leftLeftNext = root.left.left.next != null ? root.left.left.next.val : -1;
System.out.println("next of " + root.left.left.val + " is " + leftLeftNext); //4
int leftRightNext = root.left.right.next != null ? root.left.right.next.val : -1;
System.out.println("next of " + root.right.right.val + " is " + leftRightNext); //7
}
}
第四个面试官
Duy (Head of Engineering) 14:00 - 14:40
The interviewer asked me to sum (Collection<Object> objects). The objects are possible ( string, number, List, Set or HashMap). I need to check the objs’ type, and sum the values of the objects. Skip the strings of Objects.
Collection<Object>
1.245
“2.345"
List
HashMap<k, v>
Pojo {
person {
int id
String name
}
分两步做
- How to deal with List, Set, Map, refer to Check if Object is instance of String, HashMap, or HashMap[ ]
obj.instanceOf();
2). How to deal with PoJO, refer to how to get fields from a pojo dynamically
You may use java reflection. For simplicity I assume your Employee calss only contains int field. But you can use the similar rules used here for getting float, double or long value. Here is a complete code -
package com.hustbill;
import java.lang.reflect.Field;
import java.util.List;
class Person {
private int salary = 100;
private int tips = 20;
private int benefit = 25;
private int cashBack = 30;
}
public class PojoDemo {
public static void main(String[] args) throws NoSuchFieldException, IllegalAccessException {
int sum = 0;
Person Person = new Person();
Field[] allFields = Person.getClass().getDeclaredFields();
for (Field each : allFields) {
if (each.getType().toString().equals("int")) {
Field field = Person.getClass().getDeclaredField(each.getName());
field.setAccessible(true);
Object value = field.get(Person);
Integer i = (Integer) value;
sum = sum + i;
}
}
System.out.println("Sum :" + sum);
}
}
第四个是 Evance
Sr. Software Engineer 14:40 - 15:30
He asked me introduce my projects and architecture. Also, asked me to introduce infrastructure of ELK (ElasticSearch, Logstash, Kibana)
Reference
Connect nodes at same level