動機

這題的題目敘述真的不太好，夢回接手新任務，霧裡看花，用力撞牆的感覺

Problem

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 groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

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 = 1Output: 1Explanation: This is the base case.

Example 2:

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

 

Constraints:

  • 1 <= n <= 30

Sol

就是看string中的數字直到變成鱉的數字之前，

  1. 有幾個
  2. 是什麼數字

最後放到string中

class Solution:
    def countAndSay(self, n: int) -> str:
        if n == 1:
            return "1"
        else:
            s = self.countAndSay(n-1)
            chars = []
            for c in s:
                if not chars or chars[-1][0] != c:
                    chars.append([c,1])
                else:
                    chars[-1][1] += 1
            ret = []
            for p in chars:
                ret += [str(p[1]), p[0]]
            return ''.join(ret)