Apixio 服務(wù)端工程師面試題

1. Redis 和 sql數(shù)據(jù)庫 比較
2. 怎樣做到load balance
3. 分布式數(shù)據(jù)庫中數(shù)據(jù)庫同時(shí)讀寫怎么防止數(shù)據(jù)不同步
4. kafka 和 graphite 實(shí)時(shí)數(shù)據(jù)監(jiān)控是怎么做的？
   5.編程題： 從起始點(diǎn)到終點(diǎn)的路徑，附實(shí)現(xiàn)方法

/*
# Bounded region of a graph that connects 's' to 'e' defined in terms of all
# paths between 's' and 'e'. 's' can have incoming links and 'e' can have
# outgoing edges.  The graph can have cycles in it.  Each node has at most two
# outgoing edges. Additional data-structures and methods may be defined.  
# Do not add new variables in the Node.
# Node has hashCode and equals defined.
#                   +---+         +---+
#             +---->|  g |<--------+ k  +---------+
#             |     +-+-+         +---+         |
#  |   +---+  |       |             ^           |
#--+-->| s +--+       |             |           |
#      +-+-+          v             |           v         ^
#        |          +---+           |         +---+       |
#        +--------->|   |-----------+-------->| e +-------+-->
#  
                 +---+                     +---+
*/
public class Node {
  public Node left;
  public Node right;
  public int data;
}

public class BoundedGraph {
  public Node s;
  public Node e;

  public BoundedGraph(Node s, Node e) {
    this.s = s;
    this.e = e;
  }

public void printGraph() {
  Set<Node> visited = new HashSet<Node>();
  innerPrintGraph(s, e, visited);
}

private void innerPrintGraph(Node s, Node e, Set<Node> visited) {
    if (visited.contains(s)) return;
    System.out.println(s.data);
    visited.add(s);
    if (s.left != null && s != e) 
innerPrintGraph(s.left, e, visited); 
}

后記： 知識(shí)點(diǎn)還不夠熟，另外，做編程題的時(shí)候，不要老想著leetcode的解答。 自己腦袋里面有思路最關(guān)鍵。我都想到了用hashmap來標(biāo)注訪問過的節(jié)點(diǎn)。沒想到直接用HashSet來標(biāo)注節(jié)點(diǎn)。 面試官都給了很充分的提示了，我當(dāng)時(shí)腦袋停止轉(zhuǎn)動(dòng)了，還沒能搞定。面試官直接給我說了解題思路。估計(jì)是把我掛掉了，我還比較遜色。

同門小師妹都拿到Google nyc的職位了。汗顏！ 繼續(xù)努力吧！