fd

Author: liwentian

amazon/src/main/java/com/leetcode/amazon/Main.java

@@ -1,16 +1,21 @@
 package com.leetcode.amazon;
 
+import java.util.HashMap;
+import java.util.LinkedHashMap;
+
 /**
  * Created by liwentian on 2017/10/16.
  */
 
 public class Main {
 
     public static void main(String[] args) {
-        EncodeAndDecodeTinyUrl s = new EncodeAndDecodeTinyUrl();
-        String shortUrl = s.encode("http://baidu.com");
-        System.out.println(shortUrl);
+        HashMap<String, String> map = new HashMap<>();
+        map.put("one", "ones");
+        map.put("two", "twos");
 
-        System.out.println(s.decode(shortUrl));
+        for (Integer key : map.keySet()) {
+            System.out.println(key);
+        }
     }
 }

doc/系统设计/ReadME.md

@@ -0,0 +1,21 @@
+1, 如何高并发
+2，如何解决单点故障
+
+3，如何高速读
+缓存，lrucache，读写分离，多台机器，缓存命中，一致性hash
+
+4，如何高速写
+5，如何负载均衡
+6，如何海量统计
+
+
+
+数据库的垂直拆分和水平拆分
+
+中间件：
+1，RPC
+2，消息
+3，数据
+
+读多写少
+copyonwrite, readlock

ebook/tree/Tree.tex

@@ -289,10 +289,12 @@ \subsubsection{Solution}
     int leftMax = maxPathSum(root.left, left);
     int rightMax = maxPathSum(root.right, right);
     if (max != null) {
-        max[0] = max(left[0], right[0], 0) + root.val; // 此处容易错
+        // 三种情况：root, root+left, root+right
+        max[0] = max(left[0], right[0], 0) + root.val;
     }
 
     // 容易错，要考虑到所有可能的情况
+    // 左边里面的，右边里面的，光root的，左中右，左中，中右
     return max(leftMax, rightMax, root.val, left[0] + right[0] + root.val,
             left[0] + root.val, right[0] + root.val);
 }

leetcode.iml

@@ -13,7 +13,7 @@
     <content url="file://$MODULE_DIR$">
       <excludeFolder url="file://$MODULE_DIR$/.gradle" />
     </content>
-    <orderEntry type="inheritedJdk" />
+    <orderEntry type="jdk" jdkName="1.8" jdkType="JavaSDK" />
     <orderEntry type="sourceFolder" forTests="false" />
   </component>
 </module>

solution/src/main/java/com/inuker/solution/system/DesignTinyURL.java

@@ -18,17 +18,20 @@ public class DesignTinyURL {
      * 3. Mapping an identifier to an URL and its reversal - Does this problem ring a bell to you?
      *
      * 4. How do you store the URLs? Does a simple flat file database work?
-     * 采用分布式，负载均衡随机分发到一台机器，KV数据库，以短网址为key，长网址为value
+     * key为short, value为url, 则总共空间(6 + 30) * 30G = 1080G，单机恐怕不够，而且容易形成
+     * 单点故障。最好分散到多个机器上，一致性hash负载均衡，KV数据库，以短网址为key，长网址为value
+     * 每个机器上有固定前缀，后面的序号依次递增即可
      *
      * 5. What is the bottleneck of the system? Is it read-heavy or write-heavy?
      * 读显然比写多，比如用户在微博分享了一个资源，以短链接的形式。很多用户访问这个链接，则读远大于写。
-     *
+     * 所以要用cache，读写分离
      *
      * 6. Estimate the maximum number of URLs a single machine can store.
      *
      * 7. Estimate the maximum number of queries per second (QPS) for decoding a shortened URL in a single machine.
      *
      * 8. How would you scale the service? For example, a viral link which is shared in social media could result in a peak QPS at a moment's notice.
+     * 
      *
      * 9. How could you handle redundancy? i,e, if a server is down, how could you ensure the service is still operational?
      * 一致性哈希，虚拟节点