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发帖数: 1 | 3 Naval History and Heritage Command
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1902... 阅读全帖 |
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y**o
发帖数: 8897 | 4 三个挂了的都是黑帮，搜出大量枪支，通向该区的交通一度中断，司机被告知避免路经
该地区。
Assalto a restaurante termina com três bandidos mortos na Tijuca
RIO — Três pessoas morreram no assalto ao Restaurante Brasa Gourmet, na
esquina das ruas Professor Gabizo com Mariz e Barros, na Tijuca, na manh&#
227; desta segunda-feira. Segundo os policiais do 4º BPM (São
Cristóvão), as vítimas são criminosos que faziam parte da
quadrilha de assaltantes.
O comandante do 4º BPM, tenente-coronel Ronald Freitas de Santana,
disse ... 阅读全帖 |
|
c***1
发帖数: 34 | 5 你去过中国吗？当今中国怎么样？
Xiaoxiao
hi,i am chinesegirl, 21years old. as for my own county i donnt think i can
give you a veryimpersonal answer here. But i invite you to come here and
have a look, itshould be very different to your idea about China. it is not
heaven, either thehell, but we should respect the counrty that exists 5000
years yet. if you arean easygoing guy and not so sensitive with chulture
shock, you will feel verypleasant amony the Chinese people, they are very
generous and will treat youth... 阅读全帖 |
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o**********e
发帖数: 18403 | 6 我们不等待领袖的出现． 我们坚持道德制高点，但是我们必须掌握主动权。 所以
该干实事了。 星星之火，可以燎原 is really a statement about viral effect
and exponential effect.
Please come sign up to the todo list for our next step. I need only get 4-5
people to sign up. That 4-5 people get 4-5 more of their friends to sign
up. A few rounds, we are a stronger community as a result.
I created a signup sheet and please use "boycottDisney" as the accesscode:
http://www.signupgenius.com/index.cfm?go=s.SignupLogin&urlid=10
这里定目标：
http://www.mitbbs... 阅读全帖 |
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o**********e
发帖数: 18403 | 7 我们不等待领袖的出现． 我们坚持道德制高点，但是我们必须掌握主动权。 所以
该干实事了。
星星之火，可以燎原 is really a statement about viral effect and exponential
effect.
Please come sign up to the todo list for our next step. I need only get 4-5
people to sign up. That 4-5 people get 4-5 more of their friends to sign
up. A few rounds, we are a stronger community as a result.
I created a signup sheet and please use "boycottDisney" as the accesscode:
http://www.signupgenius.com/index.cfm?go=s.SignupLogin&urlid=10
这里定目标：
http://www.mitbbs... 阅读全帖 |
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T*********I
发帖数: 10729 | 8 总统正在致力于国内的income inequality 和 education inequality，其它事情先放
到下一任总统的todo list里。 |
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s****g
发帖数: 1795 | 9 Toyota corolla is a 神车 which is infamous for steering itself off the road.
Here is your todo list:
1.stop saying that is your fault
2.go to hospital now to check your neck
3.contact a lawyer
Do not contact the police to try to void your ticket. No need to contact the
dealer. Let the lawyer handle it for you. |
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r***c
发帖数: 121 | 10 谢谢大家的回复！
LG可能要求高。大概做事情也不是最高效率。
todo list上一堆一堆的事情，schedule上安排恨不得几分钟上厕所的时间也安排好，
但是我观察每天的list很少最后都完成的。
有没有这里过来人能建议要怎么做能让他focus 一点？
还有关于term insurance哪个公司的好？可以推荐一下么？ 这种家族有不好病史的
term insurance是不是会特别贵？
谢谢大家了！！ |
|
r***c
发帖数: 121 | 11 谢谢大家的回复！
LG可能要求高。大概做事情也不是最高效率。
todo list上一堆一堆的事情，schedule上安排恨不得几分钟上厕所的时间也安排好，
但是我观察每天的list很少最后都完成的。
有没有这里过来人能建议要怎么做能让他focus 一点？
还有关于term insurance哪个公司的好？可以推荐一下么？ 这种家族有不好病史的
term insurance是不是会特别贵？
谢谢大家了！！ |
|
t*********5
发帖数: 700 | 12 太对了
延伸一下第二条
如果你机器里同时运行太多的软件，再好的机器也会崩溃的。
另，给老公发todo list的习惯最让人讨厌了。 |
|
o****e
发帖数: 916 | 13 added to my todo list, will be in my next version.
most likely it won't be an auto-thing, as it needs to iterate through whole
database, and delete duplicated files. i will add an action in the tools
menu in database management. |
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a****l
发帖数: 430 | 14 You mean it will be a button? When you click it, it iterate through the
whole
database roughly twice and move/delete redundant files on hard drive and
entries from the song list. If you want a criteria, the smaller files with
identical titles should be deleted/moved :)
added to my todo list, will be in my next version.
most likely it won't be an auto-thing, as it needs to iterate through whole
database, and delete duplicated files. i will add an action in the tools
menu in database management... 阅读全帖 |
|
o****e
发帖数: 916 | 15 单屏显示的话，如果你最大化播放器，把鼠标移到屏幕的顶端，点歌节目自动会出来，
点完歌后把鼠标从点歌界面移出来就可以了，界面大小是可调的，位置目前只支持右上
角。
确实可以考虑弄一个简单的浮动窗口做单屏界面的搜索，要做的太多了，先把这个加到
TODO list吧，呵呵
界面的易用性确实是一个很大的学问，有的用户要功能强大，有的要直观简单，众口难
调，我会把那个简化版的搜索窗口（一个搜索框，加一个列表）加上，希望对不是很熟
练计算机操作的有帮助。
最近在集中时间做手机的客户端，这些应用应该会相当简洁好用 |
|
o****e
发帖数: 916 | 16 单屏显示的话，如果你最大化播放器，把鼠标移到屏幕的顶端，点歌节目自动会出来，
点完歌后把鼠标从点歌界面移出来就可以了，界面大小是可调的，位置目前只支持右上
角。
确实可以考虑弄一个简单的浮动窗口做单屏界面的搜索，要做的太多了，先把这个加到
TODO list吧，呵呵
界面的易用性确实是一个很大的学问，有的用户要功能强大，有的要直观简单，众口难
调，我会把那个简化版的搜索窗口（一个搜索框，加一个列表）加上，希望对不是很熟
练计算机操作的有帮助。
最近在集中时间做手机的客户端，这些应用应该会相当简洁好用 |
|
e********3
发帖数: 258 | 17 public class Movie{
private String name;
private List theaters;
}
public class Theater{
private List movies;
pirvate Zipcode zipcode;
}
public class Zipcode{
private Integer zipNum;
private List theaters;
}
public class MovieDaoImpl implements MovieDao {
public List findMovieByName( String name ) {
// TODO: implement this method.
}
}
public class TheaterSearchingServiceImpl implements TheaterSearchingService{
// Inject Mo |
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n********5
发帖数: 323 | 18 练习一下
public class FindMissNum {
/**
* @param args
*
* Find the missing number larger than 0.
*
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
// array elements are larger than 0.
int[] array0 = { 2, 4, 1, 9 };
int[] array1 = {1, 2,3,4};
for (int i = 0; i < array1.length; i++) {
System.out.println(array1[i]);
}
int missNum = FindNum(array1);
Syste... 阅读全帖 |
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a*********0
发帖数: 2727 | 19 泪奔，我都没调，现在work了，full code
import java.util.*;
public class SimpleRegMatch {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println("reg:?");
Scanner in=new Scanner(System.in);
String reg=in.nextLine();
System.out.println("string:?");
String input=in.nextLine();
if(StringMatch(... 阅读全帖 |
|
i*******6
发帖数: 107 | 20 #5.输入一个linkedlist和一个数字例如：9->7->8->6->1->2 和 3，输出还是一个
linkedlist但是每三个数reverse一下，例如8->7->9->2->1->6。
扩展了这道题，写成可以自己设置每几个数reverse一下。
思路大概是这样,一个原始链表为
0->1->2->3->4->5->6
如果每3个数反转一下的话，那么先把5指向0,6指向3,2指向null （换言之则是把第kn
个节点指向(k-2)*n+1个节点，再把第n个节点指向null），可以得到
6->3->4->5->0->1->2
然后把整个链表反转就是我们要的结果：
2->1->0->5->4->3->6
附上java代码和数据结构，以及测试用例，在netbeans5.5测试通过：
class ListElement{
public ListElement next;
public int data;
public ListElement(){
}
public ListElement(int data){
th... 阅读全帖 |
|
i*******6
发帖数: 107 | 21
不用啊，我没说清楚？给个代码好了：
public static int binarySearch(char[] charset, int start, int end, char target){
if (charset == null) return 0;
if (start == end){
if (charset[start] == target) return 1;
else return 0;
}
int mid = start + (end-start)/2;
if (charset[mid]
else if(charset[mid] > target) return binarySearch(charset, start, mid - 1, target);
else{
int left = 0, right = 0;
if (charset... 阅读全帖 |
|
y*****z
发帖数: 9 | 22 这个是概率题。。
先生成 4个bit的 二进制 true代表1，false代表0
这样我们能生成0-15的 等概率的 distribution
然后 [0,15] 当中 选[0,10] -->[0,2](true), [3,9](false)
public class Solution {
/**
* @param args
*/
public void doit(){
int iterate = 10000000;
boolean valve = true;
StringBuilder s = new StringBuilder();
while (valve) {
StringBuilder str = new StringBuilder();
for (int i = 0; i < 4; i++) {
if (Math.random() < 0.5) str.append('1');
... 阅读全帖 |
|
y*****z
发帖数: 9 | 23 这个是概率题。。
先生成 4个bit的 二进制 true代表1，false代表0
这样我们能生成0-15的 等概率的 distribution
然后 [0,15] 当中 选[0,10] -->[0,2](true), [3,9](false)
public class Solution {
/**
* @param args
*/
public void doit(){
int iterate = 10000000;
boolean valve = true;
StringBuilder s = new StringBuilder();
while (valve) {
StringBuilder str = new StringBuilder();
for (int i = 0; i < 4; i++) {
if (Math.random() < 0.5) str.append('1');
... 阅读全帖 |
|
m****i
发帖数: 650 | 24 It is still a solution without sorting or hashing.
public void permutationRecursive(int[] arr, int index) {
if (arr.length == index) {
for (int i = 0; i < arr.length; ++i) {
System.out.printf("%d", arr[i]);
}
System.out.println();
return;
}
int lastSwap = 0;
for (int i = index; i < arr.length; ++i) {
if (arr[i] == arr[index] && i != index) continue;
if (arr[i] == lastSwap) c... 阅读全帖 |
|
s*******f
发帖数: 1114 | 25 //回报本版，码遍本版
//Given a string of sorted integers, e.g. "1 52 69 456789 994546566";
//and a a number e.g. 69.
//You need to tell if it is in the input, e.g. 69=>true.
//strlen is O(n), don't use C style string for O(log n), suppose
//the string is friendly without lots of blank.
void GetWordPos(const char *mid, const char *left, const char *right, const
char **pstart, const char **pend){
while (isspace(*mid))
++mid;
*pstart = mid;
while (*pstart >= left && !isspace(**pstart))
... 阅读全帖 |
|
a****x
发帖数: 89 | 26 网上搜到的都没有考虑negative的情况，比如一个tree只有一个node，这个node的值
是negative。
下面是我自己的答案，已经pass OJ：
class LocalResult
{
int localMaxPathSum; // local max path sum
int localSubPathSum; // either root, or root+left, or root+right
public LocalResult(int maxPathSum, int subPathSum)
{
this.localMaxPathSum = maxPathSum;
this.localSubPathSum = subPathSum;
}
}
public class binaryTreeMaxPathSum {
/**
* @param args
*/
public static void main(String[] args) {
/... 阅读全帖 |
|
b*2
发帖数: 94 | 27 去年还真做过这道题：
private static boolean match(String regEx, String word) {
// TODO Auto-generated method stub
if(regEx == "*")
return true;
int reIndex = 0;
int wdIndex = 0;
for(;reIndex
if(regEx.charAt(reIndex)==word.charAt(wdIndex)){
continue;
}else if(regEx.charAt(reIndex)!='?'&®Ex.charAt(reIndex)!='*'){
//simply not equivalent
return false;
}else if(regEx.charAt(reIndex)=='?'){
//deal with ?
//goto the next char
continue;
}else{
//de... 阅读全帖 |
|
c****7
发帖数: 13 | 28 原题是leetcode上的merge k sorted lists
http://discuss.leetcode.com/questions/204/merge-k-sorted-lists
这个题有好几种解法，比如divide and conquer, priority queue.
我写了个priority queue的。我的问题是， 如果现在要做 merge k sorted
array,怎样来设计这个priority queue才能最简单有效呢？
附上 merge-k-sorted-lists的代码
public static ListNode mergeKLists_priorityQueue(ArrayList kLists)
{
// border case
if(kLists.size()==0) return null;
if(kLists.size()==1) return kLists.get(0);
// create a Comparator based on the nod... 阅读全帖 |
|
b*****n
发帖数: 482 | 29 是的，稍后再写个heap的，以前写过heap的class。现在为了赶进度，就先把它放到我
的Todo list里了。目前进度55/125 :) |
|
e******i
发帖数: 106 | 30 昨天做了storm8 的online code，挂了。
题目变了，不再是以前说的find max sum path in one grid。
题目如下：
给定一个string,如 “codility”,每次向左循环一个char.
codility 0th;
odilityc 1st;
dilityco 2nd;
ilitycod 3rd;
....
codility 8th;
要求返回Unique 的string. 如上所示，应当返回 7.
然后又举例，“byebye”，应当返回二
任何string，包括空数组，应当最少返回1.
要求time complexity 和 space complexity 都为O(N).
我的code：
import java.util.HashMap;
public class Cyclic {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
... 阅读全帖 |
|
e******i
发帖数: 106 | 31 昨天做了storm8 的online code，挂了。
题目变了，不再是以前说的find max sum path in one grid。
题目如下：
给定一个string,如 “codility”,每次向左循环一个char.
codility 0th;
odilityc 1st;
dilityco 2nd;
ilitycod 3rd;
....
codility 8th;
要求返回Unique 的string. 如上所示，应当返回 7.
然后又举例，“byebye”，应当返回二
任何string，包括空数组，应当最少返回1.
要求time complexity 和 space complexity 都为O(N).
我的code：
import java.util.HashMap;
public class Cyclic {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
... 阅读全帖 |
|
j**7
发帖数: 143 | 32 第三题的答案：
public class InfixToPostfix {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
String infix="4*((5+6)*7";
infix="3+(1*2)";
String postfix=toPostfix(infix);
System.out.println(postfix);
System.out.println(evaluate(postfix));
}
//evaluate a postfix equation
static int evaluate(String postfix)
{
Stack st=new Stack阅读全帖 |
|
j**7
发帖数: 143 | 33 Amazon的第四道面试题跟CodeEval上面的一道题几乎一样。可惜面试的时候没有检查到
bug.
https://www.codeeval.com/open_challenges/59/
public class Main {
/**
* @param args
*/
static StringBuilder build=new StringBuilder();
public static void main(String[] args) {
// TODO Auto-generated method stub
telephone("4155230");
}
static void telephone(String input)
{
telephone(input,0,"");
build.deleteCharAt(build.length()-1);
System.out.println(build.toString());
}
... 阅读全帖 |
|
j**7
发帖数: 143 | 34 http://www.yodlecareers.com/puzzles/triangle.html
我的答案不对 (665321)。
public static void main(String[] args) {
// TODO Auto-generated method stub
File file = new File("triangle.txt");
try
{
BufferedReader in = new BufferedReader(new FileReader(file));
String line;
int value=Integer.parseInt(in.readLine().trim());
int index=0;
while ((line = in.readLine()) != nul... 阅读全帖 |
|
j**7
发帖数: 143 | 35 结果：732506
public static void main(String[] args) {
// TODO Auto-generated method stub
File file = new File("triangle.txt");
try
{
BufferedReader in = new BufferedReader(new FileReader(file));
String line;
List list=new ArrayList();
while ((line = in.readLine()) != null) {
String [] temp=line.split("[ ]");
... 阅读全帖 |
|
s**********4
发帖数: 3 | 36 会写c++，只是拿leetcode一个例子来问一下java的函数调用问题：
比如说我现在创建了两个文件：Solution.java和SolutionTest.java:
Solution.java:
public class Solution {
public int maxProfit(int[] prices) {
// Start typing your Java solution below
// DO NOT write main() function
if (prices.length < 2)
return 0;
int maxDiff = 0;
int minValue = prices[0];
for (int i = 1; i < prices.length; ++i) {
if (prices[i] < minValue)
minValue = prices[i];
... 阅读全帖 |
|
a**c
发帖数: 52 | 37 来自主题: JobHunting版 - G家电面题 Boolean checkBasicVowel(char c) {
if (c == ‘a’ || c ==’e’ || c == ‘i’ || c ==’o’ ||
c == ‘u’ || c == ‘A’ || c ==’E’ || c == ‘I’ ||
c ==’O’ ||c == ‘U’)
return True;
}
// check whether it is vowel
Boolean checkVowel(String s, int index){
char c = s.charAt(index);
if (checkBasicVowel(c))
return True;
if (index == 0)
return False;
if (c == ‘y’ || c == ‘Y’) {
if(!checkVowel(s, index - 1))
return True;
return False;
}
return False;
}
... 阅读全帖 |
|
D*T
发帖数: 75 | 38 想了一下，这个问题好像不是那么容易。楼主的code如果末尾有个空的List会出错。我
不太懂C++所以pdu的只是半懂，不过感觉我的想法和ta估计差不多。moridin的code很
神奇，我实在看不太懂。
下面是我的code，搞ArrayList还成，搞LinkedList可能就惨了。另外remove好像是ok
的。
import java.util.*;
class ArrayPosition {
ArrayList array;
int index;
ArrayPosition(ArrayList array) {
this.array = array;
index = 0;
}
Object peekItem() {
return array.get(index);
}
Object takeItem() {
return array.get(index++);
}
}
public class DeepIteratorII implements ... 阅读全帖 |
|
D*T
发帖数: 75 | 39 想了一下，这个问题好像不是那么容易。楼主的code如果末尾有个空的List会出错。我
不太懂C++所以pdu的只是半懂，不过感觉我的想法和ta估计差不多。moridin的code很
神奇，我实在看不太懂。
下面是我的code，搞ArrayList还成，搞LinkedList可能就惨了。另外remove好像是ok
的。
import java.util.*;
class ArrayPosition {
ArrayList array;
int index;
ArrayPosition(ArrayList array) {
this.array = array;
index = 0;
}
Object peekItem() {
return array.get(index);
}
Object takeItem() {
return array.get(index++);
}
}
public class DeepIteratorII implements ... 阅读全帖 |
|
g****s
发帖数: 1755 | 40 我来一个不用统计学的Draft吧，只需要知道所有可能组合就可以推出原String：
比如google的所有可能组合一共是13种：{gge,gle,ole,ogl,goe,oge,gog,ool,ggl,oog
, gol,ooe,goo}，把这些组合都放到一个ArrayList里，然后call dfsDecode()就可以
逆推了。
private static void dfsDecode(String head, ArrayList setList,
ArrayList prosStrList) {
// TODO Auto-generated method stub
if(head.length()==6){
prosStrList.add(head);
return;
}
for(int i=0; i
if(head==""){
... 阅读全帖 |
|
e***l
发帖数: 710 | 41 import java.util.ArrayList;
import java.util.HashMap;
// TODO: rename this class to "LURCahch" for Leetcode submission
public class LRUCache2 {
private ArrayList list; // list of keys
private HashMap valueMap; // key-value map
private HashMap countMap; // appearance of each keys
in the key list
private int maxSize;
public LRUCache2(int capacity) {
maxSize = capacity;
list = new ArrayList();
valueMap... 阅读全帖 |
|
D***0
发帖数: 138 | 42 网上申请的，回复的挺快，安排了code challenge，一道题，不限时，半个小时写完了
，发过去，第二天收到了thank you but 88.不知道哪里的问题。
* Write a function that takes two parameters:
* (1) a String representing a text document and
* (2) an integer providing the number of items to return.
* Implement the function such that it returns a list of Strings ordered by
word frequency,
* the most frequently occurring word first.
* Use your best judgement to decide how words are separated.
* Your solution should run in O(n) time where n is the number of cha... 阅读全帖 |
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m*****n
发帖数: 204 | 43
The following is an O(nlgn) solution for string of size n.
It maintains a heap for eligible chars, preferring longer streams.
Ineligible chars are maintained in a candidate queue, and added back
to the working queue when their position constraint is removed.
public class MinDistance {
public String rearrange(String source, int distance) {
// TODO Check params.
// WorkingQueue of eligible streams. Longest stream at head,
// but BackupStream at tail.
PriorityQueue阅读全帖 |
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a**********0
发帖数: 422 | 44 经典的consumer和producer的例子
package jc;
import java.util.*;
public class ProCon {
// a class object to store the strings
static List myListOne = new ArrayList();
public void produce(){
synchronized(myListOne){
myListOne.add("firstNameOne");
// notice all waiting threads to stop waiting
myListOne.notifyAll();
System.out.println("Adding finished and the current myListOne.
size(): " + myListOne.size());
... 阅读全帖 |
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a**********0
发帖数: 422 | 45 继续consumer和producer的例子
package jc;
import java.util.*;
//what about there are two things to be synchronized?
public class ProCon2nd {
// a class object to store the strings
static List myListOne = new ArrayList();
static List myListTwo = new ArrayList();
// this is the thing getting synchronized
// it is static as it is the same for all instances
static Object myLock = new Object();
public void produce(){
... 阅读全帖 |
|
z*t
发帖数: 13 | 46 #include
struct point{double x, double y;};
void split(vector & v, point & line_point1, point & line_point2){
int n = v.size();
if(n==2){
line_point1 = v[0];
line_point2 = v[1];
return;
}
//get left most point
int left_index = 0;
for(int i=1;i
if(v[i].x
left_index = i;
}
//transform to angle
// angle and array index
vector > angle(n-1);
for(int i=0;i
if(i==left_index)
contin... 阅读全帖 |
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o**********e
发帖数: 18403 | 47 为什么所有老中IT都应该关注思科:
因为思科比微软,IT业更晚期. 所以思科的
所有动向都是微软,IT业的strong predictor.
发信人: longspur (longspur), 信区: SanFrancisco
标 题: 思科也来了。。
发信站: BBS 未名空间站 (Sat Aug 9 00:03:26 2014, 美东)
Cisco reports its fourth quarter earnings on Wednesday, and rumors have
begun circulating that another round of layoffs are coming, maybe announced
next week, maybe in October.
https://finance.yahoo.com/news/theres-unconfirmed-rumor-big-layoff-181642109
.html
------------------------------------------
烙印思科，高通的搬运策略
http://ww... 阅读全帖 |
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e***a
发帖数: 1661 | 48 Cisco is doomed. hahaha!!! |
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o**********e
发帖数: 18403 | 49 听信"中印合璧", "中印友好"的谎言...
现在HIGH TECH 富可敌国的大公司
烙印丐帮已经大批登堂入室,准备
全部搬运回国,或者全国搬运过来.
居然还有老中天真地以为,自己累死累活,
日夜赶工, 烙印丐帮会对他的辛苦能力有所
尊重顾忌. 或者自己能力无敌,有无穷的
公司等着接收老中被烙印赶出来的技术难民.
老中真是,无限精神阿Q精神, 唯一一条
抗争的路,其实还是最安全的, 就是不去探索. |
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a*********4
发帖数: 7 | 50 来自主题: JobHunting版 - 一道面试题 a java solution:
import java.util.*;
public class FindNonDupeSubTree {
private class Node{
int val;
Node left;
Node right;
Node (int val){
this.val = val;
left = null;
right = null;
}
}
public boolean FindSubTree(Node root, Set nodeSet, ArrayList<
Integer> validTreeRoots){
if (root == null)
return true;
Set sleft = new HashSet();
Set ... 阅读全帖 |
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