271. Encode and Decode Strings

最新推荐文章于 2023-01-08 07:10:23 发布

转载最新推荐文章于 2023-01-08 07:10:23 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/jxr041100/p/7885717.html

本文介绍了一种用于在网络上传输的字符串列表编解码算法。该算法将多个字符串编码为一个单一的字符串，并能够在接收端正确地解码回原始字符串列表。此方案适用于包含任意ASCII字符的数据。

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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.

class Codec {
public:

    // Encodes a list of strings to a single string.
    string encode(vector<string>& strs) {
        string res;
        for(auto str:strs) res+=to_string(str.size())+"@"+str;
        return res;
    }

    // Decodes a single string to a list of strings.
    vector<string> decode(string s) {
        vector<string>res;
        int head = 0;
        while(head<s.size())
        {
            int pos = s.find("@",head);
            int len = stoll(s.substr(head,pos-head));
            string str = s.substr(pos+1,len);
            res.push_back(str);
            head = pos+1+len;
        }
        return res;
    }
};

// Your Codec object will be instantiated and called as such:
// Codec codec;
// codec.decode(codec.encode(strs));

转载于:https://www.cnblogs.com/jxr041100/p/7885717.html