### Question

The count-and-say sequence is the sequence of integers beginning as follows:

`1, 11, 21, 1211, 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*, generate the *n*^{th} sequence.

Note: The sequence of integers will be represented as a string.

### Stats

Frequency | 2 |

Difficulty | 2 |

Adjusted Difficulty | 1 |

Time to use | ---------- |

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

### Solution

**This is a implementation question, not difficult**.

### My code

code 1

```
public class Solution {
public String countAndSay(int n) {
String num = "1";
for (int i = 1; i < n; i++) {
num = say(num);
}
return num;
}
private String say(String input) {
// 21 -> 1211
int len = input.length();
String output = "";
int left = 0;
int right = 0;
while (right < len) {
left = right;
// forward right until right pointer to a different value
// compared to that pointed by left pointer
while (right < len && input.charAt(left) == input.charAt(right)) {
right++;
}
output += String.valueOf(right - left);
output += input.charAt(left);
}
return output;
}
}
```

code 2

```
public String countAndSay(int n) {
String s = "1";
for (int i = 2; i <= n; i ++) {
char[] nums = s.toCharArray();
String newS = "";
int len = nums.length, left = 0, right = 0;
while (right < len) {
while (right < len && nums[left] == nums[right]) right ++;
newS += (right - left) + "" + nums[left];
left = right;
}
s = newS;
}
return s;
}
```