[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 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.

给字符加码再解码,先有码再无码,题目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,具体变换方法自己设计。

Java:

public String encode(List<String> strs) {
    StringBuffer out = new StringBuffer();
    for (String s : strs)
        out.append(s.replace("#", "##")).append(" # ");
    return out.toString();
}

public List<String> decode(String s) {
    List strs = new ArrayList();
    String[] array = s.split(" # ", -1);
    for (int i=0; i<array.length-1; ++i)
        strs.add(array[i].replace("##", "#"));
    return strs;
}

Java: with streaming

public String encode(List<String> strs) {
    return strs.stream()
               .map(s -> s.replace("#", "##") + " # ")
               .collect(Collectors.joining());
}

public List<String> decode(String s) {
    List strs = Stream.of(s.split(" # ", -1))
                      .map(t -> t.replace("##", "#"))
                      .collect(Collectors.toList());
    strs.remove(strs.size() - 1);
    return strs;
} 

Java:

// Encodes a list of strings to a single string.
    public String encode(List<String> strs) {
        StringBuilder output = new StringBuilder();
        for(String str : strs){
            // 对于每个子串,先把其长度放在前面,用#隔开
            output.append(String.valueOf(str.length())+"#");
            // 再把子串本身放在后面
            output.append(str);
        }
        return output.toString();
    }

    // Decodes a single string to a list of strings.
    public List<String> decode(String s) {
        List<String> res = new LinkedList<String>();
        int start = 0;
        while(start < s.length()){
            // 找到从start开始的第一个#,这个#前面是长度
            int idx = s.indexOf('#', start);
            int size = Integer.parseInt(s.substring(start, idx));
            // 根据这个长度截取子串
            res.add(s.substring(idx + 1, idx + size + 1));
            // 更新start为子串后面一个位置
            start = idx + size + 1;
        }
        return res;
    }

Java: better

public String encode(List<String> strs) {
    StringBuffer result = new StringBuffer();

    if(strs == null || strs.size() == 0)
        return result.toString();

    for(String str: strs){
        result.append(str.length());
        result.append("#");
        result.append(str);
    }

    return result.toString();
}

// Decodes a single string to a list of strings.
public List<String> decode(String s) {
    List<String> result = new ArrayList();

    if(s == null || s.length() == 0)
        return result;

    int current = 0;
    while(true){
        if(current == s.length())
            break;
        StringBuffer sb = new StringBuffer();
        while(s.charAt(current) != '#'){
            sb.append(s.charAt(current));
            current++;
        }
        int len = Integer.parseInt(sb.toString());
        int end = current + 1 + len;
        result.add(s.substring(current+1, end));
        current = end;
    }
    return result;
} 

Java: 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));  

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));

Python:

# Time:  O(n)
# Space: O(1)
class Codec:
    def encode(self, strs):
        """Encodes a list of strings to a single string.
        :type strs: List[str]
        :rtype: str
        """
        encoded_str = ""
        for s in strs:
            encoded_str += "%0*x" % (8, len(s)) + s
        return encoded_str


    def decode(self, s):
        """Decodes a single string to a list of strings.
        :type s: str
        :rtype: List[str]
        """
        i = 0
        strs = []
        while i < len(s):
            l = int(s[i:i+8], 16)
            strs.append(s[i+8:i+8+l])
            i += 8+l
        return strs

C++:

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 = atoi(s.substr(i, found).c_str());
            res.push_back(s.substr(found + 1, len));
            i = found + len + 1;
        }
        return res;
    }
};

C++:

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 = atoi(s.substr(0, found).c_str());
            s = s.substr(found + 1);
            res.push_back(s.substr(0, len));
            s = s.substr(len);
        }
        return res;
    }
};

  

  

All LeetCode Questions List 题目汇总

原文地址:https://www.cnblogs.com/lightwindy/p/9758222.html