cc150-1.5

gistfile1.java

package ch1;
/*
1.5 Implement a method to perform basic string compression using the counts
of repeated characters. For example, the string aabcccccaaa would become
a2blc5a3. If the "compressed" string would not become smaller than the original
string, your method should return the original string.

* When we want to add something to a string constantly and without knowing the total length,
* StringBuilder is a good choice.
* We check whether the adjacent elements are same. if so, we increase count number; if not, we output the count number,
* new character and reset the count number to 1. 
* Please be aware that at the end, we cannot detect the different adjacent elements, but we still need to add the count number.
 
 Time Complexity O(n) Space O(n)
 
*/
public class CompressString {
	
	public static String compress(String arg) {
		StringBuilder newString= new StringBuilder();
		
		//i is pointer for given string
		int count=1;
		newString.append(arg.charAt(0));
		for(int i=1; i<arg.length(); i++) {
			if(arg.charAt(i) == arg.charAt(i-1)) {
				count++;
			}else {
				newString.append(count);
				newString.append(arg.charAt(i));
				count=1;
			}
			if(i==arg.length()-1) {
				newString.append(count);
			}
		}
		
		if(newString.length()>=arg.length()){
			return arg;
		}else {
			return newString.toString();
		}	
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		String test = "aabcccccaaa";
		System.out.println(compress(test));

	}

}