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.

这道题让我们给字符加码再解码，先有码再无码，然后题目中并没有限制加码的方法，那么只要能成功的把有码变成无码就行了，具体变换方法自己可以设计。由于需要把一个字符串集变成一个字符串，然后把这个字符串再还原成原来的字符串集，最开始博主想能不能在每一个字符串中间加个空格把它们连起来，然后再按空格来隔开，但是这种方法的问题是原来的一个字符串中如果含有空格，那么还原的时候就会被分隔成两个字符串，所以必须还要加上长度的信息，加码方法是长度 + "/" + 字符串，比如对于 "a","ab","abc"，就变成 "1/a2/ab3/abc"，那么解码的时候就有规律可寻，先寻找 "/"，然后之前的就是要取出的字符个数，从 "/" 后取出相应个数即可，以此类推直至没有 "/"了，这样就得到高清无码的字符串集了，参见代码如下：

解法一：

class Codec {
public:
    // Encodes a list of strings to a single string.
    string encode(vector<string>& strs) {
        string res = "";
        for (auto a : strs) {
            res.append(to_string(a.size())).append("/").append(a);
        }
        return res;
    }
    // Decodes a single string to a list of strings.
    vector<string> decode(string s) {
        vector<string> res;
        int i = 0;
        while (i < s.size()) {
            auto found = s.find("/", i);
            int len = stoi(s.substr(i, found - i));
            res.push_back(s.substr(found + 1, len));
            i = found + len + 1;
        }
        return res;
    }
};

上面的方法是用一个变量i来记录当前遍历到的位置，我们也可以通过修改修改s，将已经解码的字符串删掉，最终s变为空的时候停止循环，参见代码如下：

解法二：

class Codec {
public:
    // Encodes a list of strings to a single string.
    string encode(vector<string>& strs) {
        string res = "";
        for (auto a : strs) {
            res.append(to_string(a.size())).append("/").append(a);
        }
        return res;
    }
    // Decodes a single string to a list of strings.
    vector<string> decode(string s) {
        vector<string> res;
        while (!s.empty()) {
            int found = s.find("/");
            int len = stoi(s.substr(0, found));
            s = s.substr(found + 1);
            res.push_back(s.substr(0, len));
            s = s.substr(len);
        }
        return res;
    }
};

我们还可以使用更简单的压缩方法，比如在每个字符串的后面加上换行字符 '\0'，其还属于一个字符串，这样在解码的时候，只要去查找这个换行字符就可以了，参见代码如下：

解法三：

class Codec {
public:
    // Encodes a list of strings to a single string.
    string encode(vector<string>& strs) {
        string res = "";
        for (string str : strs) res += str + '\0';
        return res;
    }
    // Decodes a single string to a list of strings.
    vector<string> decode(string s) {
        vector<string> res;
        stringstream ss(s);
        string t;
        while (getline(ss, t, '\0')) {
            res.push_back(t);
        }
        return res;
    }
};

Github 同步地址：

#271

类似题目：

Count and Say

Serialize and Deserialize Binary Tree

String Compression

Count Binary Substrings

参考资料：

https://leetcode.com/problems/encode-and-decode-strings/

https://leetcode.com/problems/encode-and-decode-strings/discuss/70412/AC-Java-Solution

https://leetcode.com/problems/encode-and-decode-strings/discuss/70452/C%2B%2B-super-clean-code-using-stringstream-and-getline()

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