持续创作，加速成长！这是我参与「掘金日新计划 · 6 月更文挑战」的第19天。

1. 题目介绍（count-and-say）

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

【Translate】： count-and-say序列是由递归公式定义的数字字符串序列

• countAndSay(1) = "1"
• countAndSay(n) is the way you would "say" the digit string from countAndSay(n-1), which is then converted into a different digit string.

【Translate】：

• countAndSay(1) = "1"
• countAndSay(n)是您将“说”来自countAndSay(n-1)的数字字符串的方式，然后将其转换为不同的数字字符串。

To determine how you "say" a digit string, split it into the minima l number of groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

【Translate】： 要确定如何“说”一个数字字符串，请将其分成最小数量的组，以便每组都是具有相同字符的连续部分。然后，说出每组字的数量，然后说出字。要将该语句转换为数字字符串，请将计数替换为一个数字，并将每个语句连接起来。

For example, the saying and conversion for digit string "3322251":

【Translate】： 例如，对数字字符串“3322251”的说和转换 Given a positive integer n, return the nth term of the count-and-say sequence.

【Translate】： 给定一个正整数n，返回 count-and-say序列的第n项。

【测试用例】：

【约束】：

2. 题解

2.1 递归拼接

  ayyild1z 提供的题解 Clean Short Java Solution，通过递归来拼接获得完整的字符串。

// for the nth number, you just need to count characters of the (n-1)th number,
// for the (n-1)th number, you just need to count characters of  the (n-2)th number,
// ...

public String countAndSay(int n) {
	if(n == 1) return "1";

	StringBuilder res = new StringBuilder();

	// recursively call for (n-1) th number, "0" is only for the edge case at the end of the loop with `s.charAt(i+1)`
	String s = countAndSay(n-1) + "0"; 

	for(int i=0, c=1; i < s.length()-1; i++, c++){
		// if next digit is different, then append the count so far `c` and the digit itself, then set count `c` to zero
		if(s.charAt(i+1) != s.charAt(i)){
			res.append(c).append(s.charAt(i));
			c = 0;  // 计数清零
		}
	}

	return res.toString();
}

2.2 循环拼接字符串

  zhibzhang 提供的题解 Straightforward Java Solution。从1开始，一步步完成字符串的拼接，直到n。

    public String countAndSay(int n) {
        String s = "1";
        for(int i = 1; i < n; i++){
            s = countIdx(s);
        }
        return s;
    }
    
    public String countIdx(String s){
        StringBuilder sb = new StringBuilder();
        char c = s.charAt(0);  // 取第一个元素做比较，而不是使用i和i+1比较，可以有效防止数组越界
        int count = 1;
        for(int i = 1; i < s.length(); i++){
            if(s.charAt(i) == c){
                count++;
            }
            else
            {
                sb.append(count);
                sb.append(c);
                c = s.charAt(i);  // 更换下一个比较对象
                count = 1;
            }
        }
        sb.append(count);
        sb.append(c);
        return sb.toString();
    }