# 【LeetCode】 38. Count and Say

## Description
> The count-and-say sequence is the sequence of integers with the first five terms as following:
> 1. 1
> 2. 11
> 3. 21
> 4. 1211
> 5. 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 where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.

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

> "數說數列"是一個數列，它的的前五項為：
> 1. 1
> 2. 11
> 3. 21
> 4. 1211
> 5. 111221
> 1 讀作"一個一"，所以表示為 11
> 11 讀作"兩個一"，所以表示為 21
> 21 讀作"一個二然後一個一"，所以表示為 1211

> 給予一個整數n，且1 ≤ n ≤ 30，請產生第n個數說數列的項。

> 注意：每一項回傳的型態請轉為字串。

## Example:

Example 1:

Input: 1
Output: "1"


Example 2:

Input: 4
Output: "1211"

## Solution

* 這一題難在要看懂題目...
* 去做字串分析，觀察連續相同的字元有幾個，然後拼接成新的字串即可。

### Code

class Solution {
public:
    
    string nextStr(string s)
    {
        string ans = "";
        char c = s.front();
        int count=1;
        for(int i=1;i<s.size();i++)
        {
            if(c==s[i])
                count++;
            else
            {
                ans+=(char)(count+'0');
                ans+=c;
                
                c=s[i];
                count = 1;
            }
        }
        ans+=(char)(count+'0');
        ans+=c;
        return ans;
    }
    
    string countAndSay(int n) {
        string ans="1";
        for(int i=1;i<n;i++)
            ans=nextStr(ans);
        return ans;
    }
};

###### tags: `LeetCode` `C++`