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# 38. Count and Say
## 1.問題
## 2.想法
* 提問
* 時間複雜度與可用空間是否有限制?
* 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
* 測試計算複雜度
## **3.程式碼**
```
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 cache(n, "");
return count(n, cache);
}
private:
string count(int n, vector& 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];
}
};
```
---
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