leetcode271- Encode and Decode Strings- medium

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.

本题难度不在对字符加密，而在怎么把各个字符串分开。本来字符已经占用了任意可用的256种了，所以你不能简单地选一个分隔符然后用split。

 

法1. 加密时用 长度+'#'+真正字符串的形式去加（实践真的可行，哪怕实际有分隔符的字符，因为你说明了长度了你把它取出来后下一个使用indexOf就不会影响了。而且第一个分隔符肯定来自于真正的分隔符，所以递推下去就都不会有问题了）。

解密时每次 1.indexOf('#', idx)，从指针所指的位置开始找第一个分隔符。2.上个指针到当前分割点语意理解后就是字符串长度。3.后面取出字符串即可。

 

法2. 参考ascii字符\ 转义字符的区别方法，用两个字符合在一起表示分隔符。”:;”为分隔符，并且对原字符串里所有的’:'字符做转换，变为”::”。这样就能明确地知道，序列化后字符串里遇到一个’:’，到底是分隔符的’:’呢，还是普通文本的’:’呢。（还是担心不保险可以跑一跑ab: cd, ab:; cd，都没问题的因为就算原来有:;也会被改成::;，而decode的时候看到:会先把后面的:一起吃掉，就只剩下;了，不会触发分隔。）

encode时：看到’:’转为”::”，看到结尾加个”:;”

decode时：看到’:’看下一个，若是’:’只加一个并且跳一格，若是’;’加入ans清空sb

细节：清空sb可以delete(0, length())或者赋值新的sb，并不自带clear()方法。 

法1：

public class Codec {

    // Encodes a list of strings to a single string.
    public String encode(List<String> strs) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < strs.size(); i++) {
            sb.append(strs.get(i).length());
            sb.append('#');
            sb.append(strs.get(i));
        }
        return sb.toString();
    }

    // Decodes a single string to a list of strings.
    public List<String> decode(String s) {
        List<String> result = new ArrayList<>();
        int idx = 0;
        while (idx < s.length()) {
            int cut = s.indexOf('#', idx);
            int length = Integer.parseInt(s.substring(idx, cut));
            result.add(s.substring(cut + 1, cut + 1 + length));
            idx = cut + length + 1;
        }
        return result;
    }
}

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

法2：

public class Codec {

    // Encodes a list of strings to a single string.
    public String encode(List<String> strs) {
        StringBuilder sb = new StringBuilder();
        for (String str : strs) {
            for (char c : str.toCharArray()) {
                if (c == ':') {
                    sb.append("::");
                } else {
                    sb.append(c);
                }
            }
            sb.append(":;");
        }
        return sb.toString();
    }

    // Decodes a single string to a list of strings.
    public List<String> decode(String s) {
        List<String> ans = new ArrayList<>();
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) != ':') {
                sb.append(s.charAt(i));
            } else if (i + 1 < s.length() && s.charAt(i + 1) == ':') {
                sb.append(':');
                i++;
            } else {
                ans.add(sb.toString());
                sb.delete(0, sb.length());
                i++;
            }
        }
        return ans;
    }
}

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