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# Problem 38: Count and Say

> <https://leetcode.com/problems/count-and-say/>

## 思路

* 理解题目： &#x20;

  解释一下就是，输入n，那么我就打出第n行的字符串。 &#x20;

  怎么确定第n行字符串呢？他的这个是有规律的。 &#x20;

  n = 1时，打印一个1   &#x20;

  n = 2时，看n=1那一行，念：1个1，所以打印：11   &#x20;

  n = 3时，看n=2那一行，念：2个1，所以打印：21   &#x20;

  n = 4时，看n=3那一行，念：一个2一个1，所以打印：1211   &#x20;

  以此类推。(注意这里n是从1开始的）
* 每个 entry 的格式是： count + oldChar；所以我们对每个 char 遍历一遍就好

```java
public class Solution {
    public String countAndSay(int n) {
        String oldString = "1";
        //n = n - 1;
        while (--n > 0) {
            StringBuilder sb = new StringBuilder();
            char[] oldChars = oldString.toCharArray();

            for (int i = 0; i < oldChars.length; i++) {
                int count = 1;
                while (i + 1 < oldChars.length && oldChars[i] == oldChars[i + 1]) {
                    count++;
                    i++;
                }
                sb.append(String.valueOf(count) + String.valueOf(oldChars[i]));
            }
            oldString = sb.toString();
            //n--;
        }
        return oldString;
    }
}
```

## 易错点

1. oldString = "1" 初始时已经有了，所以在 n 内已经占用了一个了。所以我们从 n - 1 开始
2. String to char

   ```java
   char[] oldChars = oldString.toCharArray();
   ```
3. int or Char to String

   ```java
   String a = String.valueOf(count);
   String b = String.valueOf(oldChars[i]);
   ```