# 38. Count and Say

## 1.問題

![](https://901207480-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LGKoChvN9am4__HCIRK%2F-LS2HsnX0AIUbh_2M3jD%2F-LS2I0ZyEdF5_0YD-BZ4%2F%E8%9E%A2%E5%B9%95%E5%BF%AB%E7%85%A7%202018-11-24%20%E4%B8%8A%E5%8D%889.18.19.png?alt=media\&token=5485e7ef-2f5d-47d2-b7bb-f7d663d5b96b)

## 2.想法 <a href="#id-2-xiang-fa" id="id-2-xiang-fa"></a>

* 提問
  * 時間複雜度與可用空間是否有限制?&#x20;
  * n是否會 <= 0?
* function header, parameter
* test input
* 說明想法
  * 觀察規律
  * 可以用遞迴的方式求解
    * n == 1時, 返回"1"
    * n != 1時, 是觀察前一個string的字母而組成
    * 因此等到反彈的字串返回時, 進行count and say &#x20;
* 測試計算複雜度

## **3.程式碼** <a href="#id-3-cheng-shi" id="id-3-cheng-shi"></a>

```
class Solution {
public:
    string countAndSay(int n) {
        if (n == 1) {
            return "1";
        } 
        
        string newStr = countAndSay(n - 1);
        string res = "";
        int index = 0, cnt = 1, tmp = 0;
        int size = newStr.length();
        while (index < size - 1) {
           if (newStr[index] != newStr[index + 1]) {
               res.push_back(char(cnt + '0'));
               res.push_back(newStr[index]);
               cnt = 1;
           } else {
               cnt++;
           }
           index++;
        }
        res.push_back(char(cnt + '0'));
        res.push_back(newStr[index]);
            
        return res;
    }
};
```

```
class Solution {
public:
    string countAndSay(int n) {
        if (n == 1) {
            return "1";
        }
        string s = "1";
        for (int i = 0; i < n - 1; i++) {
            s = getString(s);
        }
        return s;
    }
private:
    string getString(string s) {
        string res = "";
        int count = 1;
        for (int i = 0; i < s.length() - 1; i++) {
            if (s[i] != s[i + 1]) {
                res.push_back(char(count + '0'));
                res.push_back(s[i]);
                count = 1;
            } else {
                ++count;
            }
        }
        res.push_back(char(count + '0'));
        res.push_back(s[s.length() - 1]);
        
        return res;
    }
};
```

```
class Solution {
public:
    string countAndSay(int n) {
        vector<string> cache(n, "");
        return count(n, cache);
    }
private:
    string count(int n, vector<string>& cache) {
        if (n == 1) {
            return "1";
        } else if (n == 2) {
            return "11";
        }
        if (cache[n - 1] == "") {
            string s = count(n - 1, cache);
            string res = "";
            char tmp = s[0];
            int cnt = 1;
            for (int i = 1; i < s.length(); i++) {
                if (s[i] == tmp) {
                    cnt++;
                } else {
                    res = res + to_string(cnt);
                    res.push_back(tmp);
                    tmp = s[i];
                    cnt = 1;
                }
            }
            res = res + to_string(cnt);
            res.push_back(tmp);
            cache[n - 1] = res;
        }
        return cache[n - 1];
    }
};
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://jenhsuan.gitbook.io/algorithm/leetcode/38.-count-and-say.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.