Basic String Compression Input: aabbc Output: a2b2c1 CC 150 - Arrays and Strings

Compression.java

package leetcode;

import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;

/**
 * aabcccccaaa  --> a2b1c5a3
 * PS: like aa repeat is continious... Actually don't need to Use HashMap to statistic -- like book first
 * 
 * 
 * Still two ways to solve this problem. more efficient
 */

public class Compression {
	public static void main(String args[]) {
		String str = "aabcccccaaa";
		System.out.println(compression(str));
	}
	
	/**
	 * Book frist Solution O(n+k2)
	 * Same as Two pointer, if char changed, stop and append the char the repeat number 
	 */
	
	public static String compression(String str) {
		StringBuffer sb = new StringBuffer();
		char last = str.charAt(0);
		int count = 0;
		for (int i=0; i< str.length(); i++) {
			if (str.charAt(i) == last) {
				count++;
			} else {
				sb.append(last + "" + count);
				count = 1;  // here count should be 1 but not 0
			}
			last = str.charAt(i);
		}
		sb.append(last + "" + count); //reflush the last part
		
		String newStr = sb.toString();
		if (newStr.length() < str.length()) {
			return newStr;
		} else {
			return str;
		}
	}
	
	
	/**
	 * My Solution: Store every number frequency and then Reconstruct them use a StringBuffer
	 * not efficient because of two loop
	 */
	public static String compression2(String str) {
		Map<Character, Integer> map = new LinkedHashMap<Character, Integer>();
		for (int i=0; i<str.length(); i++) {
			char value = str.charAt(i);
			if(map.containsKey(value)) {
				int count = map.get(value) + 1;
				map.put(value, count );
			} else {
				map.put(value, 1);
			}
		}
		
		// construct string
		StringBuffer sb = new StringBuffer();
		Set<Entry<Character, Integer>> entrySet = map.entrySet();
		Iterator<Entry<Character, Integer>> iterator = entrySet.iterator();
		while (iterator.hasNext()) {
			Entry<Character, Integer> entry = iterator.next();
			sb.append(entry.getKey());
			sb.append(entry.getValue());
		}
		
		String newStr = sb.toString();
		// exclude situation like abc --> a1b1c1  
		if (newStr.length() < str.length()) {
			return newStr;
		} else {
			return str;
		}	
	}
	
}

import java.util.*;

public class HelloWorld{

 public static void main(String []args){
    System.out.println("Hello World");
    String str = "aabbbcccc";
    compare(str);
 }

  public static void compare(String str) {
    
        Map<Character,Integer> stringMap = new HashMap<Character,Integer>();
        
        for(int i = 0 ; i < str.length(); i++) {
            char value = str.charAt(i);
            if (stringMap.containsKey(value)) {
                int count = stringMap.get(value)+1;
                stringMap.put(value,count);
            }else{
                stringMap.put(value,1);
            }
        }
        
        String res = "";
        for (Map.Entry<Character,Integer> entry : stringMap.entrySet()) {
            res = res + ""+entry.getKey() + entry.getValue();
            //System.out.println(""+entry.getKey() + entry.getValue()); 
        } 
        System.out.println(res);
    }

}

String str= "aabbcccaaaa";

	char[] c= str.toCharArray();
	char temp=str.charAt(0);
	int count=1;
	StringBuffer sb= new StringBuffer();
	for(int i=0;i<c.length-1;i++) {
		temp=c[i];
			if(temp==c[i+1]) {
				count++;
			}else {
				sb.append(temp+""+count);
				count=1;
			}
		
		
	
		
	}
	sb.append(temp+""+count);
	System.out.println(sb.toString());

o/p: a2b2c3a4