题目：
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
evalor serialize methods. You should implement your own encode/decode algorithm.
链接： http://leetcode.com/problems/encode-and-decode-strings/
题解：
encode and decode。这里我们可以维护一个StringBuilder，读出每个input string的长度，append一个特殊字符，例如'/'，再append string。这样再decode的时候我们就可以利用java的String.indexOf(char，startIndex)来算出自startIndex其第一个'/'的位置，同时计算出接下来读取的string长度，用String.substring()读出字符串以后我们更新index，来进行下一次读取。 这些只是简单地encode/decode，至于加密之类的还需要学习Cousera上的Crytography I和II， 作业很难，希望下次开课能坚持下去。
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));
二刷:
这回也写得比较快。
在encode时我们可以对strs先append长度，再append一个delimiter，最后append目标字符串。
在decode时我们从头遍历String s，先保存一个sliding window的左端点lo，遇到delimiter的时候，我们回头去找这个字符串的长度，也就是s.substring(lo, i)。之后我们按照这个长度，把字符串extract出来，并且加入到结果集里，再更新lo以及i。最后返回结果就可以了。
稍快一点的方法可能是在decode时把字符串转换为数组然后处理，但原理大都一致。
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));
Reference:
https://leetcode.com/discuss/55020/ac-java-solution
https://leetcode.com/discuss/59840/clean-code-standard-way-of-serialization-deserialization
https://leetcode.com/discuss/57890/1-7-lines-python-length-prefixes
https://leetcode.com/discuss/54906/accepted-simple-c-solution