Compress and decompress string

You are given a string with lower case letters only. Compress it by putting the count of the letter after it. If the letter appears once, 

Example:

compress function:

input = aaaabbc output = a4b2c

decompress function:

input = a2bc3 output = aabccc

 1 public class Solution {
 2     public String compress(String input) {
 3         int startIndex = 0;
 4         StringBuilder output = new StringBuilder();
 5         while (startIndex < input.length()) {
 6             int nextIndex = findNextIndex(input, startIndex);
 7             int length = nextIndex - startIndex;
 8             output.append(input.charAt(startIndex));
 9             if (length != 1) {
10                 output.append(length);
11             }
12             startIndex = nextIndex;
13         }
14         return output.toString();
15     }
16 
17     // a4b2c
18     public String decompress(String input) {
19         int startIndex = 0;
20         StringBuilder output = new StringBuilder();
21         while (startIndex < input.length()) {
22             char ch = input.charAt(startIndex);
23             int numberEndIndex = findNumberEndIndex(input, startIndex);
24             if (numberEndIndex - startIndex == 1) {
25                 repeatLetters(output, 1, ch);
26             } else {
27                 int count = Integer.parseInt(input.substring(startIndex + 1, numberEndIndex));
28                 repeatLetters(output, count, ch);
29             }
30             startIndex = numberEndIndex;
31         }
32         return output.toString();
33     }
34 
35     private int findNumberEndIndex(String input, int startIndex) {
36         int endIndex = startIndex + 1;
37         while (endIndex < input.length() && isNumber(input.charAt(endIndex))) {
38             endIndex++;
39         }
40         return endIndex;
41     }
42 
43     private boolean isNumber(char ch) {
44         return ch >= '0' && ch <= '9';
45     }
46 
47     private void repeatLetters(StringBuilder output, int count, char ch) {
48         while (count >= 1) {
49             output.append(ch);
50             count--;
51         }
52     }
53 
54     private int findNextIndex(String input, int startIndex) {
55         int endIndex = startIndex + 1;
56         while (endIndex < input.length() && input.charAt(endIndex) == input.charAt(startIndex)) {
57             endIndex++;
58         }
59         return endIndex;
60     }
61 }

follow-up

如果字符串里有数字，比如44443333aaabbc, 它会被转成 /44/34a3b2c

其实对于这种情况的处理还是比较简单的，我们只需要知道当前letter是否是数字，如果是数字，我们就在前面加‘/’。对于decompress, 我们也需要判断当前是否是‘/’，如果是，表明第一个是数字，数字后面的到下一个‘/’之前都是个数。

 1 public class Solution {
 2     public String compress(String input) {
 3         int startIndex = 0;
 4         StringBuilder output = new StringBuilder();
 5         while (startIndex < input.length()) {
 6             int nextIndex = findNextIndex(input, startIndex);
 7             int length = nextIndex - startIndex;
 8             char letter = input.charAt(startIndex);
 9             if (isNumber(letter)) {
10                 output.append('/');
11             }
12             output.append(letter);
13             if (length != 1) {
14                 output.append(length);
15             }
16             startIndex = nextIndex;
17         }
18         return output.toString();
19     }
20 
21     // /44/35b11a4c
22     // b11a4c/44/35
23     public String decompress(String input) {
24         int startIndex = 0;
25         StringBuilder output = new StringBuilder();
26         while (startIndex < input.length()) {
27             char ch = input.charAt(startIndex);
28             if (ch == '/') {
29                 startIndex++;
30                 ch = input.charAt(startIndex);
31             }
32             int numberEndIndex = findNumberEndIndex(input, startIndex);
33             if (numberEndIndex - startIndex == 1) {
34                 repeatLetters(output, 1, ch);
35             } else {
36                 int count = Integer.parseInt(input.substring(startIndex + 1, numberEndIndex));
37                 repeatLetters(output, count, ch);
38             }
39             startIndex = numberEndIndex;
40         }
41         return output.toString();
42     }
43 
44     private int findNumberEndIndex(String input, int startIndex) {
45         int endIndex = startIndex + 1;
46         while (endIndex < input.length() && isNumber(input.charAt(endIndex))) {
47             endIndex++;
48         }
49         return endIndex;
50     }
51 
52     private boolean isNumber(char ch) {
53         return ch >= '0' && ch <= '9';
54     }
55 
56     private void repeatLetters(StringBuilder output, int count, char ch) {
57         while (count >= 1) {
58             output.append(ch);
59             count--;
60         }
61     }
62 
63     private int findNextIndex(String input, int startIndex) {
64         int endIndex = startIndex + 1;
65         while (endIndex < input.length() && input.charAt(endIndex) == input.charAt(startIndex)) {
66             endIndex++;
67         }
68         return endIndex;
69     }
70 }