Question
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.
Solution 1 -- JSON format
第一种方法是参考的JSON的规则。
Encode: 我们将输入的字符串数组封装成JSON中的array。[ (left bracket) and ] (right bracket) 表示开头的结尾。中间的分隔用, (comma)。然后对于每个字符串，是 wrapped in double quotes。
由于字符串中本来就可能有双引号或是back slash (\)，所以我们需要对这两种符号做转义。方法是多加一个back slash
如 原字符串 \"aafg" -> \\\"aafg\"
JSON里还有更复杂的字符串处理方法。但我们这里的目标只是让encode，再decode后的字符串相同，所以不必那么复杂。
Decode处理原则如下
1. 一个boolean的variable记录当前应该是下一个字符串的开头还是当前字符串的结束
2. 碰到bracket，根据是开始／结束，新建一个空字符串／将当前的字符串存入结果中
3. 碰到back slash，看它下一个元素是否是back slash / bracket，如果是，则将它下一个元素加到字符串中，计数加一。
1 public class Codec {
2 private final char start = ‘[‘;
3 private final char end = ‘]‘;
4 private final char include = ‘"‘;
5 private final char strSplit = ‘,‘;
6
7 // Encodes a list of strings to a single string.
8 public String encode(List<String> strs) {
9 StringBuilder sb = new StringBuilder();
10 sb.append(start);
11 for (String str : strs) {
12 sb.append(include);
13 int len = str.length();
14 for (int i = 0; i < len; i++) {
15 char current = str.charAt(i);
16 if (current == ‘"‘ || current == ‘\\‘) {
17 sb.append(‘\\‘);
18 }
19 sb.append(current);
20 }
21 sb.append(include);
22 sb.append(strSplit);
23 }
24 sb.append(end);
25 return sb.toString();
26 }
27
28 // Decodes a single string to a list of strings.
29 public List<String> decode(String s) {
30 List<String> result = new ArrayList<String>();
31 if (s == null || s.length() < 1) {
32 return result;
33 }
34 int len = s.length();
35 if (s.charAt(0) != start || s.charAt(len - 1) != end) {
36 return result;
37 }
38 boolean startSymbol = true;
39 StringBuilder sb = new StringBuilder();
40 for (int i = 1; i < len - 1; i++) {
41 char current = s.charAt(i);
42 if (current == include) {
43 if (startSymbol) {
44 sb = new StringBuilder();
45 } else {
46 result.add(sb.toString());
47 }
48 startSymbol = !startSymbol;
49 continue;
50 }
51 if (current == strSplit && startSymbol) {
52 continue;
53 }
54 if (current == ‘\\‘) {
55 char next = s.charAt(i + 1);
56 if (next == ‘\\‘ || next == ‘"‘) {
57 sb.append(next);
58 i++;
59 continue;
60 }
61 }
62 sb.append(current);
63 }
64 return result;
65 }
66 }
67
68 // Your Codec object will be instantiated and called as such:
69 // Codec codec = new Codec();
70 // codec.decode(codec.encode(strs));
Solution 2
利用了Java里String的 int indexOf(int ch, int fromIndex)函数。
同时存入字符串和字符串的长度。
1 public class Codec {
2
3 // Encodes a list of strings to a single string.
4 public String encode(List<String> strs) {
5 StringBuilder sb = new StringBuilder();
6 for (String str : strs) {
7 sb.append(str.length()).append(‘/‘).append(str);
8 }
9 return sb.toString();
10 }
11
12 // Decodes a single string to a list of strings.
13 public List<String> decode(String s) {
14 List<String> result = new ArrayList<String>();
15 int length = s.length();
16 int i = 0;
17 while (i < length) {
18 int slash = s.indexOf(‘/‘, i);
19 int size = Integer.valueOf(s.substring(i, slash));
20 result.add(s.substring(slash + 1, slash + size + 1));
21 i = slash + size + 1;
22 }
23 return result;
24 }
25 }