Leetcode
2020.04.25 15:54

# 38. Count and Say

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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. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.

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

Example 1:

```Input: 1
Output: "1"
Explanation: This is the base case.
```

Example 2:

```Input: 4
Output: "1211"
Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".```

```class Solution {
public String countAndSay(int n) {
if(n <= 0){
return null;
}

String result = "1";

for(int i=1; i<n; i++){
//String str = "";
StringBuilder sb = new StringBuilder();
int count = 1;

for(int j=1; j<result.length(); j++){
if(result.charAt(j) == result.charAt(j - 1)){
count++;
}else{
//str = str + (count + "" + (result.charAt(j - 1)));
sb.append(count);
sb.append(result.charAt(j - 1));
count = 1;
}
}

//str = str + (count + "" + result.charAt(result.length() - 1));
//result = str;
sb.append(count);
sb.append(result.charAt(result.length() - 1));
result = sb.toString();
}

return result;
}
}```

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