/*
 * CC150 1.5 Compress String
 * @author: Scarlett Chen
 * @time: 12/20/2014 Sat 9:57 PM
 * 时空复杂度都是o(n)?如果n是字符串里的字符数的话
 * 思路是：
 *  遍历每个字符，是否与前一个一样，一样就把count+1.不一样就把该字符和它的count放到压缩后的字符串内。
 *  当newString长度大于originalString时，继续对比就没有意义了，直接返回原来的字符串。
 *
 *
 * 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.
 */
public class CompressString {
	public String compress(String s) {
		int len = s.length();
		if (len<=2) return s;
		char[] ch = s.toCharArray();
		StringBuilder newString = new StringBuilder();
		char currentChar=' ';
		int currentCount = 0;
		for (int i=0; i<len; i++) {
			if (i>0 && ch[i]==ch[i-1]) {
				currentCount++;
			}
			else {
				if (i>0) {
					newString.append(currentChar);
					newString.append(currentCount);
					if (newString.length()>len) return s;
				}
				currentChar = ch[i];
				currentCount = 1;

			}
		}
		if (newString.length()>= len) return s;
		else return newString.toString();
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		CompressString cs = new CompressString();
		String s = "aabcccccaaa";
		System.out.println(s+"--->"+cs.compress(s));
	     s = "aa";
		System.out.println(s+"--->"+cs.compress(s));
		 s = "#";
			System.out.println(s+"--->"+cs.compress(s));
			s = "Jiali Chen";
			System.out.println(s+"--->"+cs.compress(s));

	}

}