# 38.Count and Say

**38.Count and Say**

难度:Easy

> 报数序列是一个整数序列，按照其中的整数的顺序进行报数，得到下一个数。其前五项如下：

1. ```
   1
   ```
2. ```
   11
   ```
3. ```
   21
   ```
4. ```
   1211
   ```
5. ```
   111221
   ```

1 被读作 "one 1" ("一个一") , 即 11。 11 被读作 "two 1s" ("两个一"）, 即 21。 21 被读作 "one 2", "one 1" （"一个二" , "一个一") , 即 1211。

给定一个正整数 n（1 ≤ n ≤ 30），输出报数序列的第 n 项。

注意：整数顺序将表示为一个字符串。

```
示例 1:

输入: 1
输出: "1"
示例 2:

输入: 4
输出: "1211"
```

和斐波那契数列很像，可以采用同样的方法。

```
class Solution {
private:
    vector<string> say={"1","11","21","1211"};
public:
    string countAndSay(int n) {
        if(n<=say.size()) return  say[n-1];
        string tmp=countAndSay(n-1);
        string res;
        int k=1;
        for(int i=1;i<tmp.length();i++)
        {
            if(tmp[i]==tmp[i-1])
                k++;
            else
            {
                res+=to_string(k)+tmp[i-1];
                k=1;
            }
        }
        res+=to_string(k)+tmp[tmp.length()-1];
        say.push_back(res);
        return res;
        
    }
};
```

> 执行用时 : 12 ms, 在Count and Say的C++提交中击败了70.33% 的用户\
> 内存消耗 : 9.8 MB, 在Count and Say的C++提交中击败了20.77% 的用户


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