## Question

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

`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.

To determine how you "say" a digit string, split it into the **minimal** number of substrings such that each substring contains exactly **one** unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.

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

:

Given a positive integer `n`

, return *the* `nth`

*term of the **count-and-say** sequence*.

**Example 1:**

```
Input: n = 1
Output: "1"
Explanation: This is the base case.
```

**Example 2:**

```
Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"
```

**Constraints:**

`1 <= n <= 30`

## Algorithm

Nothing special, implementation. Just do what the question title say, count and say.

You count the number and keep the current count number and its count, then if the next number changes, we put the count and number into the result string.

We do such thing from 1 to n to get the target nth count and say.

## Code

class Solution { public String countAndSay(int n) { String s = "1"; int index = 2; while (index <= n) { char ch = s.charAt(0); int count = 1; StringBuilder temp = new StringBuilder(); for (int i = 1; i < s.length(); i++) { if (s.charAt(i - 1) == s.charAt(i)) { count++; } else { temp.append(count).append(ch - '0'); ch = s.charAt(i); count = 1; } } temp.append(count).append(ch - '0'); s = temp.toString(); index++; } return s; } }