LeetCode 38. Count and Say 报数（Java）

题目：

The count-and-say sequence is the sequence of integers with the first five terms as following:

Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.

Note:
Each term of the sequence of integers will be represented as a string.

解答：

用递归的思想，直接循环推出即可
先递归调用countAndSay(n-1),之后对获取到的字符串进行解析，循环这个字符串的长度，每次与上一个字符判断是否相同，相同，则count++,否则在结果中插入count和pre，然后重置count和pre。

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