Encode & Decode

浏览数：36 / 时间：2015年06月09日

最近被各种 encode 和 decode 算法虐得很晕，总结一下，但愿有通解??

No.1 Compress a string to remove duplicates when a character occurs +2 times continuously.
e.g.
input : "12 abbbbbbbbbbbbbbc 3aaaa12bbxxx2 hello"
output: "12 a14xbc 1x34xa12bb3xx2 hello".

分析 Corner Case
* 1. number + repeat character : 3aaaaa -> 1x35xa
* 2. "number"+"x"+"character" already exist in original string : 3xa -> 1x31xxa

 1 public class CompressString {
 2     public static void main(String[] args) {
 3 //        String str = "12 abbbbbbbbbbbbbbc 3aaaa12bb3x223444 hello";
 4         String str = "123456c2xc";
 5         if (str == null || str.length() == 0)
 6             return;
 7         StringBuilder output = new StringBuilder();
 8         output.append("");
 9         int i = str.length()-1, j = str.length()-1;
10         int cur_num = 0;
11         boolean prex = false;
12         char cur = str.charAt(j);
13         while (i >= 0) {
14             if (str.charAt(i) == cur) {
15                 i--;
16             } else {
17                 cur_num = j - i;
18                 if (cur >= ‘0‘ && cur <=‘9‘) {
19                     if (!prex && cur_num <= 2) {
20                         int c = 0;
21                         while (c < cur_num) {
22                             output.insert(0, cur);
23                             c++;
24                         }
25                         prex = false;
26                     } else {
27                         output.insert(0, cur);
28                         output.insert(0, ‘x‘);
29                         output.insert(0, cur_num);
30                         prex = true;
31                     }
32                 } else if (cur_num <= 2 && cur!=‘x‘) {
33                     int c = 0;
34                     while (c < cur_num) {
35                         output.insert(0, cur);
36                         c++;
37                     }
38                     prex = false;
39                 } else {
40                     output.insert(0, cur);
41                     output.insert(0, ‘x‘);
42                     output.insert(0, cur_num);
43                     prex = true;
44                 }
45                 cur = str.charAt(i);
46                 j = i;
47                 i--;
48             }
49         }
50         cur_num = j - i;
51         if (cur >= ‘0‘ && cur <=‘9‘) {
52             if (!prex && cur_num <= 2) {
53                 int c = 0;
54                 while (c < cur_num) {
55                     output.insert(0, cur);
56                     c++;
57                 }
58             } else {
59                 output.insert(0, cur);
60                 output.insert(0, ‘x‘);
61                 output.insert(0, cur_num);
62             }
63         } else if (cur_num <= 2) {
64             int c = 0;
65             while (c < cur_num) {
66                 output.insert(0, cur);
67                 c++;
68             }
69         } else {
70             output.insert(0, cur);
71             output.insert(0, ‘x‘);
72             output.insert(0, cur_num);
73         }
74         System.out.println(output.toString());
75     }
76 }

View Code

No.2 There is a decode method F, which could
F ("someString" + "x" + "anyNumber") --> "someString""someString"..."someString" (n times of such a string)
e.g.
aaax -> ax3x

分析 Corner Case
* 1. repeat string + number : aaa3 -> ax2a3
* 2. "string"+"x"+"number" exists in the original input string : ax32aaa ->

这个暂时还有些疑惑

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Encode & Decode