213. String Compression【easy】

Implement a method to perform basic string compression using the counts of repeated characters. For example, the string aabcccccaaa would become a2b1c5a3.

If the "compressed" string would not become smaller than the original string, your method should return the original string.

You can assume the string has only upper and lower case letters (a-z).

Have you met this question in a real interview?

Example

str=aabcccccaaa return a2b1c5a3
str=aabbcc return aabbcc
str=aaaa return a4

解法一：

 public class Solution {

     /**

      * @param str a string

      * @return a compressed string

      */

     public String compress(String str) {

         // Corner case

         if (str == null) {

             return null;

         } else if (str.length() <= 2) {

             return str;

         }

         StringBuilder sb = new StringBuilder();

         char temp = str.charAt(0);

         int count = 1;

         for (int i = 1; i < str.length(); i++) {

            // If same char continues, then add the count

            if (str.charAt(i) == temp) {

                count++;

            } else {

            // Encounter different char, set temp to the char and count to 1

               sb.append(temp);

               sb.append(count);

               temp = str.charAt(i);

               count = 1;

            }

        }                  

       // Do not forget the last char and the count!!!

       sb.append(temp).append(count);                  

       // Compare the result of original str with the new stringbuilder

       if (sb.length() >= str.length()) {

             return str;

       } else {

             return sb.toString();

       }

    }

 }

Here we use StringBuilder to create a new String, instead of String concatenation. That’s because for String concatenation operation, it’ll build a new string for every operation.

The basic algorithm is:
1、Create stringbuilder and initialize the first temp char, with count = 1
2、Go through the string char by char
（1）If char(i) equals to temp, continue and add count
（2）If not, add the temp and count to stringbuilder, reset the temp to be char(i) and count to be 1
3、Go out of the loop and add the last char and count for it
4、Compare the stringbuilder with initial str

Note: After go through the for loop, the temp is equals the last - 1 char, so we need to add the last char with its count.

Time complexity: O(n)
Space complexity: O(n)

参考@Steven Wu 的代码

https://codebysteven.wordpress.com/2016/03/14/lintcode-string-compression/

解法二：

 public class Solution {

     /*

      * @param str: a string

      * @return: a compressed string

      */

     public String compress(String str) {

         if (str == null || str.length() < 3) {

             return str;

         }

         StringBuilder sb = new StringBuilder();

         Map<Character, Integer> map = new HashMap<>();

         char pre = str.charAt(0);

         map.put(pre, 1);

         for (int i = 1; i < str.length(); i++) {

             char cur = str.charAt(i);

             if (cur == pre) {

                 map.put(pre, map.get(pre)+1);

             } else {

                 sb.append(pre);

                 sb.append(map.get(pre));

                 map.put(pre, 0);

                 map.put(cur, 1);

                 pre = cur;

             }

         }

         sb.append(pre);

         sb.append(map.get(pre));

         String res = sb.toString();

         return res.length() < str.length() ? res : str;

     }

 }

参考@linspiration 的代码

https://segmentfault.com/a/1190000012634492

巴特西

213. String Compression【easy】

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