443. String Compression

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.




// correct

public int compress(char[] chars) {
        int indexAns = 0, index = 0;
        while(index < chars.length){
            char currentChar = chars[index];
            int count = 0;
            while(index < chars.length && chars[index] == currentChar){
                index++;
                count++;
            }
            chars[indexAns++] = currentChar;
            if(count != 1)
                for(char c : Integer.toString(count).toCharArray()) 
                    chars[indexAns++] = c;
        }
        return indexAns;
    }





// wrong result

任何问题 都要想清楚再 implement


class Solution {
    public int compress(char[] chars) {
        // two pointers, fast , slow 
        // things left of slow is the result, fast traverse 
        // if fast = fast + 1 , fast ++ , until fast stops at certains point, coutn = fast - slow
        // if count == 1, slow doesn't change, and slow 
        // if count > 1, slow ++, slow = count, 
        // return slow at the end 
        int slow = 0;
        int fast = 0;
        while(fast < chars.length){ // is it okay to have two while loops ? 
            while(fast + 1 < chars.length && chars[fast]== chars[fast + 1]){
                fast++;
            }
            int count = fast - slow;
            if (count > 1){
                chars[slow++] = chars[fast];
                chars[slow++] = (char) (count + '0');
            }else{
                chars[slow++] = chars[fast];
            }
            fast++; 
        }
        return slow;
    }
}


// int a = 1;
//char b = (char)(a + '0');
// Input:
// ["a","a","b","b","c","c","c"]
// Output:
// ["b","\u0003","c","\u0004"]
// Expected:
// ["a","2","b","2","c","3"]