# 38. Count and Say

## 13 Jan 38. Count and Say

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

• `countAndSay(1) = "1"`
• `countAndSay(n)` is the way you would “say” the digit string from `countAndSay(n-1)`, which is then converted into a different digit string.

To determine how you “say” a digit string, split it into the minimal number of substrings such that each substring contains exactly one unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.

For example, the saying and conversion for digit string `"3322251"`:

Given a positive integer `n`, return the `nth` term of the count-and-say sequence.

Example 1:

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

Example 2:

```Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"
```

Constraints:

• `1 <= n <= 30`

Java 代码如下：

public class Solution {
public String countAndSay(int n) {
StringBuilder ans = new StringBuilder(“1”);
StringBuilder prev;
int count;
char say;
for(int i = 1; i < n; i++){
prev = ans;
ans = new StringBuilder();
count = 1;
say = prev.charAt(0);
for(int j = 1; j < prev.length(); j++){
if(say != prev.charAt(j)){
ans.append(count).append(say);
count = 1;
say = prev.charAt(j);
}else{
count++;
}
}
ans.append(count).append(say);
}
return ans.toString();
}
}

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python 代码如下：

class Solution(object):
def countAndSay(self, n):
“””
:type n: int
:rtype: str
“””
res = “1”
for i in range(n – 1):
prev = res[0]
count = 1
ans = “”
for j in range(1, len(res)):
cur = res[j]
if prev != cur:
ans = ans + str(count) + str(prev)
prev = cur
count = 0
count += 1
res = ans + str(count) + str(prev)
return res

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countAndSay(1) = “1”
countAndSay(n) 是对 countAndSay(n-1) 的描述，然后转换成另一个数字字符串。

C++ 代码如下：

class Solution {
public:
string countAndSay(int n) {
if (n == 1) {
return “1”;
}
string before = countAndSay(n – 1);
string res;
char cur = before[0];
int count = 1;
for (int i = 1; i < before.size(); ++i) {
if (before[i] != cur) {
res += to_string(count) + cur;
cur = before[i];
count = 0;
}
count ++;
}
res += to_string(count) + cur;
return res;
}
};
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class Solution:
def countAndSay(self, n: int) -> str:
results = “1”
for i in range(n-1):
previous = results[0]