G面经prepare: Friends Recommendation

想想如果你用linkedin或者facebook, 给你一个人和他的朋友关系网,你会怎么给一个人推荐朋友一个例子就是A-B, A-C, B - D, B - E, C - D,这个时候问我应该推荐谁给A,我说D,因为他是BC的共同好友,而E只是B的好友,到这我才明白干啥,就是给一个图和里面的一个节点A,用bfs从A出发,找出第二层中indegree度数最大节点用HashMap<Character, HashSet<Character>>来建图用HashMap<Chara…

G面经prepare: Set Intersection && Set Difference

求两个sorted数组的intersection e.g. [1,2,3,4,5],[2,4,6] 结果是[2,4] difference 类似merge, 分小于等于大于三种情况,然后时间O(m+n), 空间O(1) package ShortestSubsequenceIncluding; import java.util.*; public class Solution2 { public ArrayList<Integer> setInters(int[] arr1, int[] ar…

G面经prepare: Reorder String to make duplicates not consecutive

字符串重新排列,让里面不能有相同字母在一起.比如aaabbb非法的,要让它变成ababab.给一种即可 Greedy: 跟FB面经Prepare task Schedule II很像,记录每个char出现次数,然后用最大堆,把剩下char里面出现次数多的优先Poll出来组建新的string 如果poll出来的char跟上一个相同,则用一个queue暂时存一下我觉得时间复杂度:O(N) + O(KlogK) + O(NlogK) = O(NlogK) ,where K is the number…

FB面经Prepare: Friends Recommendation

有个getFriend() API, 让你推荐你的朋友的朋友做你的朋友,当然这个新朋友不能是你原来的老朋友 package fb; import java.util.*; public class ReferFriends { public List<String> recommendation(String name) { List<String> res = new ArrayList<String>(); if (name==null || name.length…

G面经prepare: Maximum Subsequence in Another String's Order

求string str1中含有string str2 order的 subsequence 的最小长度 DP做法:dp[i][j]定义为pattern对应到i位置,string对应到j位置时,shortest substring的长度,Int_Max表示不存在 package ShortestSubsequenceIncluding; public class Solution { public String findShortest(String a, String b){ if(a==nul…

G面经prepare: Pattern Match

设定一个pattern 把 'internationalization' 变成 'i18n', 比如word是house,pattern可以是h3e, 3se, 5, 1o1s1等, 给pattern和word,判断是否match, package DataStreamAverage; public class Solution3 { public boolean check(String s1, String s2) { return helper(s1, 0, s2, 0); } publi…

G面经prepare: Data Stream Average

给一个datastream和一个fixed window size, 让我design一个class可以完成add number还有find average in the window. 就是不能用vector得用array. 当然用queue是最好的 package DataStreamAverage; import java.util.*; public class Solution { int count; int sum; Queue<Integer> q; public Soluti…

G面经prepare: Android Phone Unlock Pattern

1 2 3 4 5 6 7 8 9 只有中间没有其他键的两个键才能相连,比如1可以连 2 4 5 6 8 但不能连 3 7 9 但是如果中间键被使用了,那就可以连,比如5已经被使用了,那1就可以连9 每个键只能用一次,给定一个长度L,求问有多少unique path with length L Backtracking: 我的code不光可以知道数目,还可以打印所有Pattern package AndroidUnlockPattern; import java.util.*; public c…

G面经prepare: Jump Game Return to Original Place

第二题算法给你一个arr 返回 T 或者 F arr的每个数代表从这个点开始跳几部,返回T的情况:从这个arr中任意一个数开始跳,可以在每个元素都跳到且只跳到一次的情况下返回到开始跳的元素比如[1,1,1,1,1,1] => T [0,1,1,1,1,1]=> F [7, 5, 2, 3] => F [2,2,3,1] => T. From 1poi scan array once, get the index that can be reached by each arra…

G面经prepare: Sort String Based On Another

Given a sorting order string, sort the input string based on the given sorting order string. Ex sorting order string -> dfbcae Input string -> abcdeeabc output -> dbbccaaee 法一:Comparable sample Input: String order = "dfbcae"; String str…

G面经Prepare: Valid Preorder traversal serialized String

求问下各位大神,怎么判断一个按照Preorder traversal serialized的binary tree的序列是否正确呢?不能deserialize成树比如 A) 9 3 4 # # 1 # # 2 # 6 # #是对的,因为表示 9 / \ 3 2 / \ \ 4 1 6 B ) 9 3 4 # # 1 # #就是错的,因为无法反构造回一棵树我觉得可以在字符串里找"n##"这种结构(对应tree里两个children都是Null的叶节点),找到之后就把"n##&…

G面经prepare: Chucked Palindrome

给定一个字符串,找出最多有多少个chunked palindrome, 正常的palindrome是abccba, chunked palindrome的定义是:比如volvo, 可以把vo划分在一起,(vo) (l) (vo),那么它是个palindrome.求实现返回最大的chunk 数量. 比如aaaaaa可以是(aaa)(aaa), 但是最大chunk数量应该是(a)(a)(a)(a)(a)(a)为6 这就是一个greedy的问题,从string的两边开始,用i和j记录当前scan到的位…

G面经prepare: BuyGoods

给你一部分钱和一些不同价钱的商品,如何在最多买K件商品的情况下尽可能多的花掉手里的钱. 举例:口袋里的钱数: 10; K=2 产品价格: [3, 6, 8, 7, 9] 输出 3, 7 Backtracking: package BuyGoods; import java.util.*; public class Solution { static int minRemain = 0; public ArrayList<Integer> optimize(int money, int[] pr…

G面经prepare: X-Straight

Define “X-Straight” as X cards with consecutive numbers (X >= 3). Determine if the deck can be fully divided into sets of “X-Straight”. Example: 1, 2, 3, 4, 4, 5, 6 -> True Backtracking: package Straight; import java.util.*; public class Solution2 {…

G面经prepare: set difference

给你A{1,2,3,4,4,5}, B{2,4},求A-B={1,3,4,5},很简单. visit 1 只用一个HashMap package TwoSets; import java.util.*; public class Solution { public ArrayList<Integer> findDiff(int[] arr1, int[] arr2) { ArrayList<Integer> res = new ArrayList<Integer>();…

G面经prepare: Straight Partition of A Deck of Cards

Define “Straight” as 5 cards with consecutive numbers. Determine if the deck can be fully divided into sets of “Straight”. Example: 1, 2, 3, 4, 4, 5, 5, 6, 7, 8 -> TrueYou may assume the cards are sorted 这个是用一个hashtable,key是数字,value是出现次数然后遍历原数组,每一个数…

G面经Prepare: Search word delete sequence in dictionary

给一个单词一个字典,每次删除单词里任一个字母直到剩下一个字母,形成一个序列,比如office->offce->ofce->ofc->oc->c.问是否字典里存在一个这种序列 package checkDictExistSequence; import java.util.*; public class Solution { HashSet<String> dict = new HashSet<String>(); public String check…

G面经Prepare: Longest All One Substring

give a string, all 1 or 0, we can flip a 0 to 1, find the longest 1 substring after the flipping 这是一个简单版本of LC 424 Longest Repeating Character Replacement 又是Window, 又是Two Pointers Window还是采用每次都try to update left使得window valid, 每次都检查最大window package G…

G面经Prepare: Print Zigzag Matrix

For instance, give row = 4, col = 5, print matrix in zigzag order like: [1, 8, 9, 16, 17] [2, 7, 10, 15, 18] [3, 6, 11, 14, 19] [4, 5, 12, 13, 20] package GooglePhone; import java.util.Arrays; public class PrintMatrix { static void print(int rows, in…

Project Management Process

Project Management ProcessDescription .......................................................................................................................................................................................1STAGE/STEP/TASK SUMMARY LIST…

不容易系列之(4)——考新郎[HDU2049]

不容易系列之(4)——考新郎 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17001 Accepted Submission(s): 6348 Problem Description国庆期间,省城HZ刚刚举行了一场盛大的集体婚礼,为了使婚礼进行的丰富一些,司仪临时想出了有一个有意思的节目,叫做"考新郎",具体的操作是这…

神、上帝以及老天爷[HDU2048]

神.上帝以及老天爷 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17381 Accepted Submission(s): 7409 Problem DescriptionHDU 2006'10 ACM contest的颁奖晚会隆重开始了!为了活跃气氛,组织者举行了一个别开生面.奖品丰厚的抽奖活动,这个活动的具体要求是这样的:…

【BZOJ4555】求和（多种解法混合版本）

[BZOJ4555]求和(多种解法混合版本) 题面 BZOJ 给定\(n\),求 \[f(n)=\sum_{i=0}^{n}\sum_{j=0}^{i}S(i,j)\times 2^j \times (j!)\] \(n<=100000\),结果对\(998244353\)取模. 其中\(S(i,j)\)是第二类斯特林数,表示将\(i\)个有区别的球放入\(j\)个相同的盒子中,每个盒子非空的方案数. \(S(n,m)=S(n-1,m-1)+S(n-1,m)*m\) 边界条件:\(S(i,0)=…

近年Recsys论文

2015年~2017年SIGIR,SIGKDD,ICML三大会议的Recsys论文: [转载请注明出处:https://www.cnblogs.com/shenxiaolin/p/8321722.html] SIGIR-2015 [Title]WEMAREC: Accurate and Scalable Recommendation through Weighted and Ensemble Matrix Approximation [Abstract]Matrix approximation…

RTP Payload Format for Transport of MPEG-4 Elementary Streams over http

1．SDP (1)Http Request GET /getSdpForUrl?HttpUrl=nphMpeg4/g726-640x480 HTTP/1.0/r/n Host: 58.63.71.90:8011/r/n Accept: */*/r/n Accept-Language: */r/n Accept-Encoding: */r/n Authorization: Basic NjU0MzIxOjEyMzQ1NgA=/r/n User-Agent: Streaming Sdk 1.0/r/…