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

 public class Solution {
     public String compress(String input) {
         int startIndex = ;
         StringBuilder output = new StringBuilder();
         while (startIndex < input.length()) {
             int nextIndex = findNextIndex(input, startIndex);
             int length = nextIndex - startIndex;
             output.append(input.charAt(startIndex));
             if (length != ) {
                 output.append(length);
             }
             startIndex = nextIndex;
         }
         return output.toString();
     }

     // a4b2c
     public String decompress(String input) {
         int startIndex = ;
         StringBuilder output = new StringBuilder();
         while (startIndex < input.length()) {
             char ch = input.charAt(startIndex);
             int numberEndIndex = findNumberEndIndex(input, startIndex);
             if (numberEndIndex - startIndex == ) {
                 repeatLetters(output, , ch);
             } else {
                 int count = Integer.parseInt(input.substring(startIndex + , numberEndIndex));
                 repeatLetters(output, count, ch);
             }
             startIndex = numberEndIndex;
         }
         return output.toString();
     }

     private int findNumberEndIndex(String input, int startIndex) {
         int endIndex = startIndex + ;
         while (endIndex < input.length() && isNumber(input.charAt(endIndex))) {
             endIndex++;
         }
         return endIndex;
     }

     private boolean isNumber(char ch) {
         return ch >= '' && ch <= '';
     }

     private void repeatLetters(StringBuilder output, int count, char ch) {
         while (count >= ) {
             output.append(ch);
             count--;
         }
     }

     private int findNextIndex(String input, int startIndex) {
         int endIndex = startIndex + ;
         while (endIndex < input.length() && input.charAt(endIndex) == input.charAt(startIndex)) {
             endIndex++;
         }
         return endIndex;
     }
 }

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。

 public class Solution {
     public String compress(String input) {
         int startIndex = ;
         StringBuilder output = new StringBuilder();
         while (startIndex < input.length()) {
             int nextIndex = findNextIndex(input, startIndex);
             int length = nextIndex - startIndex;
             char letter = input.charAt(startIndex);
             if (isNumber(letter)) {
                 output.append('/');
             }
             output.append(letter);
             if (length != ) {
                 output.append(length);
             }
             startIndex = nextIndex;
         }
         return output.toString();
     }

     // /44/35b11a4c
     // b11a4c/44/35
     public String decompress(String input) {
         int startIndex = ;
         StringBuilder output = new StringBuilder();
         while (startIndex < input.length()) {
             char ch = input.charAt(startIndex);
             if (ch == '/') {
                 startIndex++;
                 ch = input.charAt(startIndex);
             }
             int numberEndIndex = findNumberEndIndex(input, startIndex);
             if (numberEndIndex - startIndex == ) {
                 repeatLetters(output, , ch);
             } else {
                 int count = Integer.parseInt(input.substring(startIndex + , numberEndIndex));
                 repeatLetters(output, count, ch);
             }
             startIndex = numberEndIndex;
         }
         return output.toString();
     }

     private int findNumberEndIndex(String input, int startIndex) {
         int endIndex = startIndex + ;
         while (endIndex < input.length() && isNumber(input.charAt(endIndex))) {
             endIndex++;
         }
         return endIndex;
     }

     private boolean isNumber(char ch) {
         return ch >= '' && ch <= '';
     }

     private void repeatLetters(StringBuilder output, int count, char ch) {
         while (count >= ) {
             output.append(ch);
             count--;
         }
     }

     private int findNextIndex(String input, int startIndex) {
         int endIndex = startIndex + ;
         while (endIndex < input.length() && input.charAt(endIndex) == input.charAt(startIndex)) {
             endIndex++;
         }
         return endIndex;
     }
 }