Monday, December 5, 2016

Interview Misc Part 3


http://ninefu.github.io/blog/RelationCosts/
 public List<Node> findShortestPath(List<List<Integer>> relations, int start, int target) {
  if (relations == null || relations.size() < 0) {
   return null;
  }
  HashMap<Integer, Node> map = new HashMap<>();
  for (int i = 0; i < relations.size(); i++) {
   Node person;
   if (map.containsKey(i)) {
    person = map.get(i);
   } else {
    person = new Node(i);
    map.put(i, person);
   }
   if (i == start) {
    person.minCost = 0;
    person.minPath.add(person);
   }
   for (Integer integer : relations.get(i)) {
    Node after = map.getOrDefault(integer, new Node(integer));
    map.put(integer, after);
    person.frontier.add(after);
   }
  }
  
  dijkstra(map, start, target);
  return map.get(target).minPath;
 }
 
 public void dijkstra(HashMap<Integer, Node> map, int start, int target) {
  PriorityQueue<Node> frontiers = new PriorityQueue<>((a,b) -> (a.minCost - b.minCost));
  Node startNode = map.get(start);
  for (Node next : startNode.frontier) {
   next.minCost = (next.person - startNode.person) * (next.person - startNode.person);
   next.minPath = new LinkedList<>(startNode.minPath);
   next.minPath.add(next);
   frontiers.add(next);
  }
  while (!frontiers.isEmpty()) {
   Node cur = frontiers.poll();
   for (Node next : cur.frontier) {
    int oldCost = next.minCost;
    int cost = cur.minCost + (next.person - cur.person) * (next.person - cur.person);
    if (cost < next.minCost) {
     next.minCost = cost;
     next.minPath = new LinkedList<>(cur.minPath);
     next.minPath.add(next);
    }
    if (oldCost == Integer.MAX_VALUE) {
     frontiers.add(next);
    }
   }
  }
 }
 
 public class Node {
  int person;
  int minCost;
  List<Node> minPath;
  List<Node> frontier;
  
  public Node(int person) {
   this.person = person;
   minCost = Integer.MAX_VALUE;
   minPath = new LinkedList<>();
   frontier = new LinkedList<>();
  }
 }
http://ninefu.github.io/blog/PageDisplay/
  • Each page has 12 listings
  • Try to avoid listings with the same host ID in the same page
  • If the number of remaining listings is less than 12, allow duplicated host ID in the same page// not implemented?
 public static List<List<String>> displayPages(List<String> input) {
  LinkedList<List<String>> result = new LinkedList<>();
  if (input == null || input.size() == 0) return result;
  
  HashSet<String> visited = new HashSet<>();
  LinkedList<String> page = new LinkedList<>();
  Iterator<String> iter = input.iterator();
  
  while (iter.hasNext()) {
   String entry = iter.next();
   String host = entry.split(",")[0];
   if (visited.add(host)) {
    page.add(entry);
    iter.remove();
   }
   if (page.size() == 12 || !iter.hasNext()) {
    visited.clear();
    iter = input.iterator();
    if (page.size() == 12) {
     // sort the list
     result.add(new LinkedList<>(page));
     page = new LinkedList<>();
    }
   }
  }
  if (page.size() != 0) {
   result.add(page);
  }
  return result;
 }
 
 public static List<List<String>> displayFast(List<String> input) {
  List<List<String>> result = new LinkedList<>();
  if (input == null || input.size() == 0) return result;
  
  int notFullPageNumber = 0;
  HashMap<String, Integer> map = new HashMap<>(); // <host id, last inserted page index>
  for (int i = 0; i < input.size(); i++) {
   String entry = input.get(i);
   String host = entry.split(",")[0];
   int page = map.getOrDefault(host, -1);
   // never seen host id, insert into the first page that is not full
   if (page == -1) {
    if (result.size() <= notFullPageNumber) {
     result.add(new LinkedList<>());
    }
    result.get(notFullPageNumber).add(entry);
    map.put(host, notFullPageNumber);
    if (result.get(notFullPageNumber).size() == 12) {
     notFullPageNumber = findNextNotFullPage(notFullPageNumber, result);
    }
   } else {
    // duplicate id, insert into the first not-full page after the last inserted index
    int nextInsertPageNumber = findNextNotFullPage(page, result);
    if (nextInsertPageNumber == result.size() && 
      result.get(nextInsertPageNumber - 1).size() + input.size() - i > 12) {
     result.add(new LinkedList<>());
    } else if (nextInsertPageNumber == result.size()){
     nextInsertPageNumber--;
    }
    result.get(nextInsertPageNumber).add(entry);
    map.put(host, nextInsertPageNumber);
   }
  }
  return result;
 }
 
 public static int findNextNotFullPage(int currentPageNumber, List<List<String>> result) {
  int res = currentPageNumber + 1;
  while (res < result.size() && result.get(res).size() == 12) {
   res++;
  }
  return res;
 }
http://ninefu.github.io/blog/MeetingIntervals/
 * 每个subarray都已经sort好
举例:
[
  [[1, 3], [6, 7]],
  [[2, 4]],
  [[2, 3], [9, 12]].
]
返回
[[4, 6], [7, 9]]
N个员工,每个员工有若干个interval表示在这段时间是忙碌的。求所有员工都不忙的intervals
解法:这题最简单的方法就是把所有区间都拆成两个点,然后排序,然后扫描,每次碰到一个点如果
是左端点就把busy_employees加1,否则减1,等到每次busy_employees为0时就是一个新的区间。
这样复杂度O(MlogM),M是总共区间数。但是我当时也不知道是脑抽了还是脑子太好用了,弄了个堆的,
堆里N个元素,存每个员工的下一个端点,如果堆顶是某个员工的左端点就busy_employees加1,否则减1。
这样好处是时间复杂度低了一点点,坏处是给面试官讲了半天而且代码不是太好写……不建议这么做.
 * @author yihuifu
 *
 */
public class MeetingIntervals {

 public static void main(String[] args) {
  MeetingIntervals meetingIntervals = new MeetingIntervals();
  List<List<Interval>> intervals = new LinkedList<>();
  for (int i = 0; i < 3; i++) {
   intervals.add(new LinkedList<>());
   switch (i) {
    case 0: 
     intervals.get(i).add(meetingIntervals.new Interval(1,3));
     intervals.get(i).add(meetingIntervals.new Interval(6,7));
     break;
    case 1:
     intervals.get(i).add(meetingIntervals.new Interval(2,8));
     break;
    case 2:
     intervals.get(i).add(meetingIntervals.new Interval(2,3));
     intervals.get(i).add(meetingIntervals.new Interval(9,12));
     break;
   }
  }
  List<Interval> freeIntervals = meetingIntervals.findFreeIntervals(intervals);
  for (Interval interval : freeIntervals) {
   System.out.println(interval.toString());
  }
 }
 
 public List<Interval> findFreeIntervals(List<List<Interval>> intervals) {
  List<Interval> result = new LinkedList<>();
  if (intervals == null || intervals.size() == 0) {
   return result;
  }
  LinkedList<Interval> all = new LinkedList<>();
  for (int i = 0; i < intervals.size(); i++) {
   for (int j = 0; j < intervals.get(i).size(); j++) {
    all.add(intervals.get(i).get(j));
   }
  }
  Collections.sort(all, (a,b) -> (a.startTime - b.startTime));
  PriorityQueue<Interval> pq = new PriorityQueue<>((a,b) -> a.endTime - b.endTime);
  pq.offer(all.get(0));
  for (int i = 1; i < all.size(); i++) {
   Interval prev = pq.poll();
   Interval cur = all.get(i);
   if (cur.startTime <= prev.endTime) {
    prev.endTime = Math.max(prev.endTime, cur.endTime);
   } else {
    pq.offer(cur);
   }
   pq.offer(prev);
  }
  Interval prev = pq.poll();
  while (!pq.isEmpty()) {
   Interval cur = pq.poll();
   result.add(new Interval(prev.endTime, cur.startTime));
   prev = cur;
  }
  return result;
 }
 
 public class Interval {
  int startTime;
  int endTime;
  
  public Interval(int start, int end) {
   startTime = start;
   endTime = end;
  }
  
  @Override
  public String toString() {
   return "" + startTime + " " + endTime;
  }
 }

http://ninefu.github.io/blog/CIDR/
 * 输入一个ip,和一个count来表示一组ip, e.g.   7.7.7.7和3表示 7.7.7.7, 7.7.7.8, 7.7.7.9, 
 * 输出这组ip的CIDR形式.CIDR表达是: ip地址/掩码位数 表示一个区间
 * 比如 0.0.0.8 / 30 就是以0.0.0.8为准,前30位不能变--》0.0.0.(0000 10 00)--0.0.0.(0000 10 11) 
 * 然后给你一个起始ip,和数量。用最少的cidr表示这个区间
 * @author yihuifu
 
 public void getCIDRS(String binaryIP, int count, List<String> res) {
  while (count > 0) {
   int zero = 0, last = binaryIP.length() - 1;
   while (last >= 0 && binaryIP.charAt(last) == '0') {
    if (count - (int) Math.pow(2, zero + 1) >= 0) {
     zero++;
     last--;
    } else {
     break;
    }
   }
   String cidrIP = binaryToDecimal(binaryIP) + "/" + (32 - zero);
   System.out.println("Found one CIDR " + cidrIP);
   res.add(cidrIP);
   count -= (int) Math.pow(2, zero);
   binaryIP = getNextBinaryIP(binaryIP, zero);
  }
  
 }
 
 public String getNextBinaryIP(String prevIp, int lastZero) {
  System.out.println("prevIP " + prevIp);
  int one = 32 - 1 - lastZero;
  while (one >= 0 && prevIp.charAt(one) == '1') {
   one--;
  }
  StringBuilder res = new StringBuilder(prevIp.substring(0, one));
  res.append("1");
  while (res.length() < 32) {
   res.append("0");
  }
  return res.toString();
 }
 
 public String binaryToDecimal(String binaryIP) {
  StringBuilder sb = new StringBuilder();
  for (int i = 0; i < binaryIP.length(); i += 8) {
   String subBinary;
   if (i < 24) {
    subBinary = binaryIP.substring(i, i + 8);
   } else {
    subBinary = binaryIP.substring(i);
   }
    
   Integer sub = Integer.parseInt(subBinary, 2);
   sb.append(sub);
   if (i < 24) {
    sb.append(".");
   }
  }
  System.out.println("convert " + binaryIP + " to " + sb.toString());
  return sb.toString();
 }
 
 public String decimalToBinary(String decimalIP) {
  StringBuilder sb = new StringBuilder();
  String[] ips = decimalIP.split("\\.");
  for (int i = 0; i < ips.length; i++) {
   Integer sub = Integer.parseInt(ips[i]);
   String binarySub = Integer.toBinaryString(sub);
   int missingDigit = 8 - binarySub.length();
   String zeros = "";
   while (missingDigit > 0) {
    zeros += "0";
    missingDigit--;
   }
   sb.append(zeros + binarySub);
  }
  System.out.println("convert " + decimalIP + " to " + sb.toString());
  return sb.toString();
 }


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