这种字符串处理的题目，微软很喜欢考

思路是在递归的基础上进行字符串处理

class Solution {
    public String countAndSay(int n) {
        if(n==1) return "1";
        String str = countAndSay(n-1);
        StringBuilder sb = new StringBuilder();
        int count = 1, i = 0;
        while(i<str.length()){
            while(i+1<str.length()&&str.charAt(i)==str.charAt(i+1)){
                count++;
                i++;
            }
            sb.append(count);
            sb.append(str.charAt(i));
            i++;
            count = 1;
        }
        return sb.toString();
    }
}