题目:

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.

链接: http://leetcode.com/problems/encode-and-decode-strings/

题解:

encode and decode。这里我们可以维护一个StringBuilder,读出每个input string的长度,append一个特殊字符,例如'/',再append string。这样再decode的时候我们就可以利用java的String.indexOf(char,startIndex)来算出自startIndex其第一个'/'的位置,同时计算出接下来读取的string长度,用String.substring()读出字符串以后我们更新index,来进行下一次读取。 这些只是简单地encode/decode,至于加密之类的还需要学习Cousera上的Crytography I和II, 作业很难,希望下次开课能坚持下去。

Time Complexity - O(n), Space Complexity - O(1)

public class Codec {

    // Encodes a list of strings to a single string.
public String encode(List<String> strs) {
if(strs == null || strs.size() == 0) {
return "";
}
StringBuilder sb = new StringBuilder();
for(String s : strs) {
int len = s.length();
sb.append(len);
sb.append('/');
sb.append(s);
}
return sb.toString();
} // Decodes a single string to a list of strings.
public List<String> decode(String s) {
List<String> res = new ArrayList<>();
if(s == null ||s.length() == 0) {
return res;
}
int index = 0;
while(index < s.length()) {
int forwardSlashIndex = s.indexOf('/', index);
int len = Integer.parseInt(s.substring(index, forwardSlashIndex));
res.add(s.substring(forwardSlashIndex + 1, forwardSlashIndex + 1 + len));
index = forwardSlashIndex + 1 + len;
}
return res;
}
} // Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.decode(codec.encode(strs));

二刷:

这回也写得比较快。

在encode时我们可以对strs先append长度,再append一个delimiter,最后append目标字符串。

在decode时我们从头遍历String s,先保存一个sliding window的左端点lo,遇到delimiter的时候,我们回头去找这个字符串的长度,也就是s.substring(lo, i)。之后我们按照这个长度,把字符串extract出来,并且加入到结果集里,再更新lo以及i。最后返回结果就可以了。

稍快一点的方法可能是在decode时把字符串转换为数组然后处理,但原理大都一致。

Java:

Time Complexity - O(n), Space Complexity - O(n)

public class Codec {

    // Encodes a list of strings to a single string.
public String encode(List<String> strs) {
StringBuilder sb = new StringBuilder();
for (String s : strs) {
sb.append(s.length()).append('#').append(s);
}
return sb.toString();
} // Decodes a single string to a list of strings.
public List<String> decode(String s) {
List<String> res = new ArrayList<>();
if (s == null || s.length() == 0) return res;
for (int lo = 0, i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '#') {
int len = Integer.parseInt(s.substring(lo, i));
res.add(s.substring(i + 1, i + 1 + len));
lo = i + 1 + len;
i = i + 1 + len;
}
}
return res;
}
} // Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.decode(codec.encode(strs));

Reference:

https://leetcode.com/discuss/55020/ac-java-solution

https://leetcode.com/discuss/59840/clean-code-standard-way-of-serialization-deserialization

https://leetcode.com/discuss/57890/1-7-lines-python-length-prefixes

https://leetcode.com/discuss/54906/accepted-simple-c-solution

271. Encode and Decode Strings的更多相关文章

  1. [LeetCode#271] Encode and Decode Strings

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

  2. [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 ...

  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

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

  7. 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 ...

  8. [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 ...

  9. 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 ...

随机推荐

  1. DES,3DES,AES这三种对称密钥的区别与联系

    DES:Data Encryption Standard(数据加密标准,又美国国密局,选中的IBM的方案,密钥长度为56,标准提出是要使用64位长的密钥,但是实际中DES算法只用了64位中的56位密钥 ...

  2. 基于OpenMP的矩阵乘法实现及效率提升分析

    一.  矩阵乘法串行实现 例子选择两个1024*1024的矩阵相乘,根据矩阵乘法运算得到运算结果.其中,两个矩阵中的数为double类型,初值由随机数函数产生.代码如下: #include <i ...

  3. Java缓冲流细节

    FileOutPutStream继承OutputStream,并不提供flush()方法的重写所以无论内容多少write都会将二进制流直接传递给底层操作系统的I/O,flush无效果.而Buffere ...

  4. Shell常用操作

    1.读取配置文件中的jdbc_url参数的值($InputParamFile为待读取的目标文件绝对路径) jdbc_url=`grep "jdbc_url" $InputParam ...

  5. CSS3选择器学习笔记

    CSS选择器总结: 一.基本选择器 1.通配选择器:[  *  ]        选择文档中所以HTML元素. *{margin: 0;padding: 0;} /*选择页面中的所有元素并设置marg ...

  6. Google Guava学习笔记——基础工具类Splitter的使用

    另一项经常对字符串的操作就是根据指定的分隔符对字符串进行分隔.我们基本上会使用String.split方法: String testString = "Monday,Tuesday,,Thu ...

  7. TCP 粘包/拆包问题

    简介    TCP 是一个’流’协议,所谓流,就是没有界限的一串数据. 大家可以想想河里的流水,是连成一片的.期间并没有分界线, TCP 底层并不了解上层业务数据的具体含义 ,它会根据 TCP 缓冲区 ...

  8. 【BZOJ】【2005】【NOI2010】能量采集

    欧拉函数 玛雅,我应该先看看JZP的论文的……贾志鹏<线性筛法与积性函数>例题一 这题的做法……仔细想下可以得到:$ans=2*\sum_{a=1}^n\sum_{b=1}^m gcd(a ...

  9. 【POJ】【2891】Strange Way to Express Integers

    中国剩余定理/扩展欧几里得 题目大意:求一般模线性方程组的解(不满足模数两两互质) solution:对于两个方程 \[ \begin{cases} m \equiv r_1 \pmod {a_1} ...

  10. Android ADT中增大AVD内存后无法启动:emulator failed to allocate memory 8 (转)

    Android ADT中增大AVD内存后无法启动:emulator failed to allocate memory 8http://www.crifan.com/android_emulator_ ...