443. String Compression

Question

Given an array of characters, compress it in-place.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.

After you are done modifying the input array in-place, return the new length of the array.

Follow up: Could you solve it using only O(1) extra space?

Example 1:

Input:
["a","a","b","b","c","c","c"]

Output:
Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]

Explanation:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".

 

Example 2:

Input:
["a"]

Output:
Return 1, and the first 1 characters of the input array should be: ["a"]

Explanation:
Nothing is replaced.

 

Example 3:

Input:
["a","b","b","b","b","b","b","b","b","b","b","b","b"]

Output:
Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].

Explanation:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
Notice each digit has it's own entry in the array.

Note:

All characters have an ASCII value in [35, 126].
1 <= len(chars) <= 1000.

Solutions:

Method 1: 9m46s two pointers

  class Solution {
      public int compress(char[] chars) {
          int count = 0;
          int read = 0;
          while(read < chars.length){
              char c = chars[read];
              int check = read;
              while(check < chars.length && chars[check] == c){
                  check++;
              }
              // c appears check - read times.
              chars[count++] = c;
              if(check - read > 1){
                  String num = "" + (check - read);
                  for(int i = 0; i < num.length(); i++){
                      chars[count++] = num.charAt(i);
                  }
              }
              read = check;
          }
          return count;
      }
  }