apanda009

2017年12月2日

摘要： or this question, the key part is: what is the state of the game? Intuitively, to uniquely determine the result of any state, we need to know: A secon 阅读全文

posted @ 2017-12-02 01:49 apanda009 阅读(268) 评论(0) 推荐(0)

rankable cache class

摘要： public class RetainBestCache { private Map cache;. 鍥磋鎴戜滑@1point 3 acres private Map> rankingOfObject;.鏈枃鍘熷垱鑷�1point3acres璁哄潧 private DataSource dataSource; private int maxSiz... 阅读全文

posted @ 2017-12-02 00:14 apanda009 阅读(6292) 评论(0) 推荐(0)

two sum class

摘要：基本思路就是递归的把k sum化成k-1 sum，直到变成trivial的2sum来做。可以用两个hashset 存元素，sum ,这样就是store O(n), get(O(1)) 阅读全文

posted @ 2017-12-02 00:09 apanda009 阅读(163) 评论(0) 推荐(0)

2017年12月1日

Generate a binary tree from parent->child relationship

摘要： This was asked in LinkedIn Interview Given a list of child->parent relationships, build a binary tree out of it. All the element Ids inside the tree are unique. Example: Given the following relatio... 阅读全文

posted @ 2017-12-01 23:47 apanda009 阅读(391) 评论(0) 推荐(0)

An implementation of this using bufferedReader:

摘要： public class TextFile implements Iterable{ private BufferedReader br; public TextFile(String fileName) throws FileNotFoundException { br = new BufferedReader(new FileR... 阅读全文

posted @ 2017-12-01 22:07 apanda009 阅读(150) 评论(0) 推荐(0)

implement list use arrary

摘要： public class MyList { private static final int DEFAULT_CAPACITY = 10; private E[] elements; private int size = 0; public MyList() { elements = new E[DEFAULT_CAPACITY]; ... 阅读全文

posted @ 2017-12-01 12:49 apanda009 阅读(145) 评论(0) 推荐(0)

12. Integer to Roman

摘要： public static String intToRoman(int num) { String M[] = {"", "M", "MM", "MMM"}; String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"}; Strin 阅读全文

posted @ 2017-12-01 11:00 apanda009 阅读(93) 评论(0) 推荐(0)

Partition of a set into K subsets with equal sum

摘要： input check, 正数？ In every recursion, we at most try k groups, which means the search tree has k branches at most. And the depth of the search tree is 阅读全文

posted @ 2017-12-01 09:29 apanda009 阅读(297) 评论(0) 推荐(0)

2017年11月30日

what's problem will you face in the multiple threading tasks

摘要： The most common concurrency problem I've seen, is not realizing that a field written by one thread is not guaranteed to be seen by a different thread. 阅读全文

posted @ 2017-11-30 22:27 apanda009 阅读(158) 评论(0) 推荐(0)

Max Points on a line

摘要： Brute Force的做法，N个点两两可以构成N(N-1)/2条线，我们可以找这N(N-1)/2条线中线上点数最大值，只需对每一条线再进行一层O(N)的遍历，总共是O(N^3)。用第二种方法更好，选一个基准点，看后面每一个点跟它构成的直线，维护一个HashMap， key是跟这个点构成直线的阅读全文

posted @ 2017-11-30 22:08 apanda009 阅读(226) 评论(0) 推荐(0)

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