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tong-1324

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Snapchat的效率非常高，第一天同学帮忙投的内推，第二天就约我当周电面，内推免去了OA，杀了我一个措手不及。实习的project正进入最忙的时候，所以也还没有太多时间准备，硬着头皮就上了。
面试排在下午，一个小时时间，是个国人小哥，人很Nice。上来先自我介绍了一下，说到现在在某公司实习，面试官对我所在的这个组已经我做的项目都很感兴趣，问得非常具体。这里提一下，我暑假实习里用的主要写的语言是C#，我和面试官对这个语言做了一些讨论，聊得还挺欢，但是这为后面一个悲剧的事情留下了伏笔。然后他又在简历里挑了一个感兴趣的项目，也问的比较具体。最后问了一下为什么想加入Snapchat。没有具体看时间，但是我

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是当时时间本来就剩的不多，自己有点慌乱，没注意检查。最后发现的时候也没法跟面试官说了，觉得比较遗憾。

补充内容 (2016-8-16 04:42):
======================8.15 update==========================
上周发完帖子之后大家提醒我可以给HR发邮件说说情况，我就发了一个简单说明了一下，刚刚收到HR的邮件，给我再加了一轮电面

pingfanzhilu

贴个2 pass的解法吧（the second pass is merge）有1 pass的吗？

import java.util.*;
class Interval {
public int left;
public int right;
public Interval(int l, int r) {
this.left = l;
this.right = r;
}
}
public class Solution {
public static void main(String[] args) {
List<String[]> input = new ArrayList();
input.add(new String[]{"f1", "start", "1"});
input.add(new String[]{"f1", "start", "2"});
input.add(new String[]{"f2", "start", "4"});
input.add(new String[]{"f2", "end", "8"});
input.add(new String[]{"f1", "end", "16"});
input.add(new String[]{"f1", "end", "32"});
input.add(new String[]{"f3", "start", "64"});
input.add(new String[]{"f3", "end", "128"});
Solution sol = new Solution();
Map<String, List<Interval>> result = sol.solve(input);
for (String k : result.keySet()) {
System.out.print(k + " : ");
for (Interval itv : result.get(k)) {
System.out.print("[" + itv.left + ", " + itv.right + "]");
}
System.out.println();
}
}
public Map<String, List<Interval>> solve(List<String[]> input) {
Map<String, List<Interval>> result = new HashMap();
String[] prev = null;
for (String[] current : input) {
if (current[1].equals("start")) {
if (prev != null && prev[1].equals("start")) {
if (!result.containsKey(prev[0])) {
result.put(prev[0], new ArrayList<>());
}
result.get(prev[0]).add(new Interval(Integer.valueOf(prev[2]), Integer.valueOf(current[2])));
}
} else { // end
if (!result.containsKey(prev[0])) {
result.put(current[0], new ArrayList<>());
}
result.get(current[0]).add(new Interval(Integer.valueOf(prev[2]), Integer.valueOf(current[2])));
}
prev = current;
}
for (String key : result.keySet()) {
result.put(key, merge(result.get(key)));
}
return result;
}
private List<Interval> merge(List<Interval> list) {
Stack<Interval> stack = new Stack();
for (Interval itv : list) {
if (stack.isEmpty()) {
stack.push(itv);
} else {
if (stack.peek().right < itv.left) {
stack.push(itv);
} else {
Interval top = stack.pop();
stack.push(new Interval(Math.min(top.left, itv.left), Math.max(top.right, itv.right)));
}
}
}
List<Interval> ret = new ArrayList();
while (!stack.isEmpty()) {
ret.add(stack.pop());
}
return ret;
}
}

复制代码

eboyme

1 pass。大概写了一下。感觉这个应当假定输入总是合法的。比如f1 start 1, f2 start 2, f1 end 4这种情况不应该存在，否则CPU在时间[2,4]内的运行状态是不确定的。

import java.util.*;
class Log{
String program;
String status;
int timeStamp;
public Log(String program, String status, int timeStamp){
this.program = program;
this.status = status;
this.timeStamp = timeStamp;
}
}
public class FunctionLogParsing {
public static Map<String, List<int[]>> parsingLogFile(Log[] file){
Map<String, List<int[]>> res = new HashMap<String, List<int[]>>();
if (file == null || file.length == 0)
return res;
Log prev = file[0];
int start = prev.timeStamp;
int end = -1;
for (int i = 1; i < file.length; i++){
Log curr = file[i];
if (curr.status.equals("start")){
if (curr.program.equals(prev.program)){
if (prev.status.equals("end")){
addRes(start, end, res, prev);
}
}
else{
if (prev.status.equals("start")){
end = curr.timeStamp;
}
addRes(start, end, res, prev);
start = curr.timeStamp;
end = -1;
prev = curr;
}
}
else {
if (curr.program.equals(prev.program)){
end = curr.timeStamp;
}
else{
addRes(start, end, res, prev);
start = end;
end = -1;
prev = curr;
}
}
}
addRes(start, end, res, prev);
return res;
}
private static void addRes(int start, int end, Map<String, List<int[]>> res, Log prev){
int[] interval = {start, end};
if (!res.containsKey(prev.program)){
res.put(prev.program, new ArrayList<int[]>());
}
res.get(prev.program).add(interval);
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Log[] logs = {
new Log("f1", "start", 1),
new Log("f1", "start", 2),
new Log("f2", "start", 4),
new Log("f2", "end", 8),
new Log("f1", "end", 16),
new Log("f1", "end", 32),
new Log("f3", "start", 64),
new Log("f3", "end", 128),
};
Map<String, List<int[]>> map = parsingLogFile(logs);
for (String key: map.keySet()){
System.out.print(key+": ");
for (int[] pair: map.get(key)){
System.out.print("[" + pair[0] + "," + pair[1] + "] ");
}
System.out.println();
}
}
}

复制代码

importcoder

我也只能想到先把相同function的interval放到map里，然后依次merge。

// assuming following situation will never occur:
// f1 start 1
// f2 start 4
// f1 end 7
// f2 end 10
// meaning if there is an "end", the previous must be a "start" with same ID
public class LogParsing {
public Map<String, List<List<Integer>>> logParser(String[] log) {
Map<String, List<List<Integer>>> result = new HashMap<>();
Stack<String[]> stack = new Stack<>();
// put cpu occupied period for each function
for (String line: log) {
String[] cur = line.split("\t");
result.putIfAbsent(cur[0], new ArrayList<>());
if (stack.isEmpty()) {
stack.push(cur);
continue;
}
String[] pre = stack.peek();
int start = Integer.parseInt(pre[2]);
int end = Integer.parseInt(cur[2]);
List<Integer> interval = Arrays.asList(start, end);
result.get(pre[0]).add(interval);
if (cur[1].equals("start")) {
stack.push(cur);
} else if (cur[1].equals("end")) {
stack.pop();
if (!stack.isEmpty()) {
stack.peek()[2] = cur[2];
}
}
}
// merge occupied time lists
for (String id: result.keySet()) {
List<List<Integer>> list = mergeIntervals(result.get(id));
result.put(id, list);
}
return result;
}
private List<List<Integer>> mergeIntervals(List<List<Integer>> intervals) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> pre = intervals.get(0);
for (int i = 1; i < intervals.size(); i++) {
List<Integer> cur = intervals.get(i);
if (overlap(pre, cur)) {
List<Integer> interval = Arrays.asList(Math.min(pre.get(0), cur.get(0)), Math.max(pre.get(1), cur.get(1)));
pre = interval;
} else {
result.add(pre);
pre = cur;
}
}
result.add(pre);
return result;
}
private boolean overlap(List<Integer> pre, List<Integer> cur) {
return pre.get(1) >= cur.get(0);
}
public static void main(String[] args) {
LogParsing solution = new LogParsing();
String[] log = {"f1\tstart\t1", "f1\tstart\t2", "f2\tstart\t4", "f2\tend\t8", "f1\tend\t16", "f1\tend\t32", "f3\tstart\t64", "f3\tend\t128"};
System.out.println(solution.logParser(log));
}
public static void print(String[] log) {
for (String s: log) {
System.out.println(s);
}
}
}

复制代码

wtcupup

谢谢楼主分享，能贴个当时写的代码看看嘛？

tong-1324

wtcupup 发表于 2016-8-12 16:29
谢谢楼主分享，能贴个当时写的代码看看嘛？

页面关掉了没有存下来哦

芥末青豆

so any feedback?

cx101220012

发邮件给hr啊解释一下大不了求他再给一次店面机会

tong-1324

芥末青豆发表于 2016-8-13 01:53
so any feedback?

暂时还没有

tong-1324

cx00001 发表于 2016-8-13 02:03
发邮件给hr啊解释一下大不了求他再给一次店面机会

恩恩，要是被拒了的话准备argue一下，但是不抱太大希望:(

cx101220012

先解释呀到时候讲就比较弱了

tong-1324

cx00001 发表于 2016-8-13 04:47
先解释呀到时候讲就比较弱了

有道理！我赶紧去说一下，谢谢提醒~

sadfcbasy

求问snapchat现在就开始了吗?因为刚问了内推的说还没开始。。

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