[LeetCode#38]Count and Say

Problem:

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the nth sequence.

Note: The sequence of integers will be represented as a string.

Analysis:

The problem is easy! You just need to construct a string from a given string iteratively. 
But it include the skill in counting forward appearance!!!
"111221", How to count it each in one pass?
We know, only at "111(2)21", we know many '1' appears before it. The same for other characters!
There is a special code structure for handling such problem.
Key: We record the result of appearance when we reach the different character.
-------------------------------------------------------------------------------
key 1: the invariant require we know the count of previous charcter. We assigin count = 1, thus we could start from j = 1.

int count = 1;
for (int j = 1; j < res.length(); j++) {
    ....
}
-------------------------------------------------------------------------------
key 2: Once the current character is the same as the prvious character, we increase the count.

if (res.charAt(j) == res.charAt(j-1)) {
    count++;
} 
-------------------------------------------------------------------------------
key 3: iff the current charcter is different from the privious one. We record the appearance of the previous character.

if (res.charAt(j) != res.charAt(j-1)) {
    buffer.append(count);
    buffer.append(res.charAt(j-1));
    count = 1;
}
-------------------------------------------------------------------------------
key 4: Since we delay the record of a character's appearance at next character, we must miss the last character's record, we must add it after the for loop.

buffer.append(count);
buffer.append(res.charAt(res.length()-1));
res = buffer.toString();
-------------------------------------------------------------------------------

Solution:

public class Solution {
    public String countAndSay(int n) {
        if (n <= 0)
            throw new IllegalArgumentException("n is not in the valid range");
        String res = "1";
        for (int i = 2; i <= n; i++) {
            StringBuffer buffer = new StringBuffer();
            int count = 1;
            for (int j = 1; j < res.length(); j++) {
                if (res.charAt(j) == res.charAt(j-1)) {
                    count++;
                } else {
                    buffer.append(count);
                    buffer.append(res.charAt(j-1));
                    count = 1;
                }
            }
            buffer.append(count);
            buffer.append(res.charAt(res.length()-1));
            res = buffer.toString();
        }
        return res;
    }
}