LeetCode 38. Count and Say

原题链接在这里：https://leetcode.com/problems/count-and-say/

题目：

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.

题解：

后一个结果是在前一个基础上得出的，采用pre 和 cur替换. 

当前的读音用来做数列的下一个"1" 读成1个1，写成"11"用来作为下一个.

计算相同char的个数先append进去.

Time Complexity: O(n * cur.length()). Space: O(cur.length()).

AC Java:

class Solution {
    public String countAndSay(int n) {
        if(n <= 0){
            return "";
        }
        
        StringBuilder sb = new StringBuilder("1");
        for(int k = 1; k<n; k++){
            String pre = sb.toString();
            sb = new StringBuilder();
            int count = 1;
            for(int i = 1; i<pre.length(); i++){
                if(pre.charAt(i) == pre.charAt(i-1)){
                    count++;
                }else{
                    sb.append(count);
                    sb.append(pre.charAt(i-1));
                    count = 1;
                }
            }
            
            sb.append(count);
            sb.append(pre.charAt(pre.length()-1));
        }
        
        return sb.toString();
    }
}