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Encode and Decode Strings 解答

时间:2015-11-09 07:09:24      阅读:359      评论:0      收藏:0      [点我收藏+]

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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 eval or 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。

技术分享

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 }

 

Encode and Decode Strings 解答

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原文地址:http://www.cnblogs.com/ireneyanglan/p/4948887.html

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