Problem:

Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.

Machine 1 (sender) has the function:

string encode(vector<string> strs) {
// ... your code
return encoded_string;
}

Machine 2 (receiver) has the function:

vector<string> decode(string s) {
//... your code
return strs;
}

So Machine 1 does:

string encoded_string = encode(strs);

and Machine 2 does:

vector<string> strs2 = decode(encoded_string);

strs2 in Machine 2 should be the same as strs in Machine 1.

Implement the encode and decode methods.

Note:

  • The string may contain any possible characters out of 256 valid ascii characters. Your algorithm should be generalized enough to work on any possible characters.
  • Do not use class member/global/static variables to store states. Your encode and decode algorithms should be stateless.
  • Do not rely on any library method such as eval or serialize methods. You should implement your own encode/decode algorithm.

Analysis:

This problem needs some skills in implementation. Once you know the tricky skill underlying it, you would think how it could be so easy!
Instant idea: Can you use some special characters to separate those strings.
Nope! No matter what kind of special characters you use, it may appear in each individual string by chance! Then I have came up with the idea to use certain number of characters to record each string's information in the overall string.
However, how much prefix characters is enough? how to sepearte the information for each string out?
That's a headache problem! The genius idea: why not combinely use special character and size information. Wrap your string in following way in the encode string.
encode_string = size1:{original_string}size2:{original_string}size3:{original_string}size4:{original_string}
each original string is wrap through following way:
original_string ---> size1:{original_string} For a single block, how could we extract the orginal_string out of wraped string?
Step 1: get the start index of the block. Inital start index is 0.
-------------------------------------------------------------------
int next_start = 0; Step 2: use ":" to get the orginal_string's length.
-------------------------------------------------------------------
int split_index = s.indexOf(":", next_start);
int len = Integer.valueOf(s.substring(next_start, split_index)); Step 3: combinely use ":" and length information to extract the original string out.
-------------------------------------------------------------------
String item = s.substring(split_index+1, split_index+1+len);
ret.add(item); Step 4: update the start index for the next string.
-------------------------------------------------------------------
next_start = split_index+1+len;

Wrong Solution:

public class Codec {
// Encodes a list of strings to a single string.
public String encode(List<String> strs) {
if (strs == null)
throw new IllegalArgumentException("strs is null");
StringBuffer buffer = new StringBuffer();
for (String str : strs) {
buffer.append(str.length());
buffer.append(":");
buffer.append(str);
}
return buffer.toString();
} // Decodes a single string to a list of strings.
public List<String> decode(String s) {
List<String> ret = new ArrayList<String> ();
int next_start = 0;
int split_index = s.indexOf(":");
int len = Integer.valueOf(s.substring(next_start, split_index));
while (next_start < s.length()) {
String item = s.substring(split_index+1, split_index+1+len);
ret.add(item);
next_start = split_index+1+len;
split_index = s.indexOf(":", next_start);
len = Integer.valueOf(s.substring(next_start, split_index));
}
return ret;
}
}

Mistakes Analysis:

Last executed input:
[] Mistake Analysis:
My first implementation is complex and so ugly!!!
Since we need to do the same work for all wrapped strings, we should not allow a singly operation spill out the common block. int next_start = 0;
int split_index = s.indexOf(":"); //what if there is no string in the encoded string!!! This ugly logic incure a corner case!
int len = Integer.valueOf(s.substring(next_start, split_index));
while (next_start < s.length()) {
String item = s.substring(split_index+1, split_index+1+len);
ret.add(item);
next_start = split_index+1+len;
split_index = s.indexOf(":", next_start);
len = Integer.valueOf(s.substring(next_start, split_index));
} What's more, "while (next_start < s.length())" is great checking for cases!

Solution:

public class Codec {
// Encodes a list of strings to a single string.
public String encode(List<String> strs) {
if (strs == null)
throw new IllegalArgumentException("strs is null");
StringBuffer buffer = new StringBuffer();
for (String str : strs) {
buffer.append(str.length());
buffer.append(":");
buffer.append(str);
}
return buffer.toString();
} // Decodes a single string to a list of strings.
public List<String> decode(String s) {
List<String> ret = new ArrayList<String> ();
int next_start = 0;
while (next_start < s.length()) {
int split_index = s.indexOf(":", next_start);
int len = Integer.valueOf(s.substring(next_start, split_index));
String item = s.substring(split_index+1, split_index+1+len);
ret.add(item);
next_start = split_index+1+len;
}
return ret;
}
} // Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.decode(codec.encode(strs));

[LeetCode#271] Encode and Decode Strings的更多相关文章

  1. [LeetCode] 271. Encode and Decode Strings 加码解码字符串

    Design an algorithm to encode a list of strings to a string. The encoded string is then sent over th ...

  2. 271. Encode and Decode Strings

    题目: Design an algorithm to encode a list of strings to a string. The encoded string is then sent ove ...

  3. [LC] 271. Encode and Decode Strings

    Design an algorithm to encode a list of strings to a string. The encoded string is then sent over th ...

  4. [LeetCode] Encode and Decode Strings 加码解码字符串

    Design an algorithm to encode a list of strings to a string. The encoded string is then sent over th ...

  5. LeetCode Encode and Decode Strings

    原题链接在这里:https://leetcode.com/problems/encode-and-decode-strings/ 题目: Design an algorithm to encode a ...

  6. Encode and Decode Strings -- LeetCode

    Design an algorithm to encode a list of strings to a string. The encoded string is then sent over th ...

  7. [Swift]LeetCode271. 加码解码字符串 $ Encode and Decode Strings

    Design an algorithm to encode a list of strings to a string. The encoded string is then sent over th ...

  8. Encode and Decode Strings

    Design an algorithm to encode a list of strings to a string. The encoded string is then sent over th ...

  9. Encode and Decode Strings 解答

    Question Design an algorithm to encode a list of strings to a string. The encoded string is then sen ...

随机推荐

  1. Mybatis高级应用

    Mybatis是一个半自动的框架.相对于hibernate全自动模式,mybatis为开发人员提供了更加灵活的对sql语句操作的控制能力,有利于dba对相关的sql操作进行优化,同时也方便开发者构建复 ...

  2. spring使用aop

    基于spring-framework-4.1.7使用aop >>>>>>>>>>>>>>>>>&g ...

  3. css3 盒模型

    0,前言 在css2.1 之前,我们都熟知的两种盒模型,一种是w3c标准盒模型,另外一种是怪异模式下的盒模型.在css3之前我们一直使用的是标准盒模型,但是标准盒模型的宽度总是需要小心的去使用,稍有不 ...

  4. Winedt 7.0 Build: 20120321 永久试用方法 WinEdt 7.0 破解

    该方法,不是破解. 因为WinEdt试用版与正式版功能无异. 所以,该方法是 通过更新注册表信息,重置安装时间. 也就是重新获取31天的试用期时长. 方法如下: 1.用管理员权限打开CMD. 2.运行 ...

  5. Struts2多文件上传

    第一步:首先创建一个多文件上传的页面 <html> <head> <meta http-equiv="Content-Type" content=&q ...

  6. C++专题 - Qt是什么

    Qt是一个1991年由奇趣科技开发的跨平台C++图形用户界面应用程序开发框架.它既可以开发GUI程式,也可用于开发非GUI程式,比如控制台工具和服务器.Qt是面向对象的框架,使用特殊的代码生成扩展(称 ...

  7. 南理第八届校赛同步赛-F sequence//贪心算法&二分查找优化

    题目大意:求一个序列中不严格单调递增的子序列的最小数目(子序列之间没有交叉). 这题证明贪心法可行的时候,可以发现和求最长递减子序列的长度是同一个方法,只是思考的角度不同,具体证明并不是很清楚,这里就 ...

  8. 多重背包的入门题目HDU1171,2191,2844.

    首先,什么叫多重背包呢? 大概意思就是:一个背包有V总容量,有N种物品,其价值分别为Val1,Val2--,Val3,体积对应的是Vol1,Vol2,--,Vol3,件数对应Num1,Num2--,N ...

  9. who am i

    本原创文章属于<Linux大棚>博客,博客地址为http://roclinux.cn.文章作者为rocrocket. 为了防止某些网站的恶性转载,特在每篇文章前加入此信息,还望读者体谅. ...

  10. yum命令学习

    yum配置文件 /etc/yum.conf yum check-update检查一下有无更新 每天都要(设置定时任务todo) 1.列出所有可更新的软件清单---yum check-update 2. ...