271. Encode and Decode Strings

 

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:

https://github.com/grandyang/leetcode/issues/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()

 

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation