LeetCode 38. Count and Say-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_40044912/article/details/127399139

本文解析了count-and-say序列的生成过程，通过递归方式描述如何将数字字符串转化为新的计数序列，如1到1211的转变。重点介绍了算法实现思路，包括通过遍历和计数来构造新序列。

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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.

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”

Solution:
countAndSay(1) = “1” doesn’t need compute, the rest repeats n-1 times.

My Code:

class Solution:
    def countAndSay(self, n: int) -> str:
        current_str = "1"
        if n == 1: return current_str
        for i in range(0,n-1):
            set_i = []
            count = 1
            current_char = current_str[0]
            for x in range(0,len(current_str)):
                if x == (len(current_str)-1):
                    set_i.append(str(count)+current_char)
                    break
                if current_str[x+1] == current_char:
                    count = count + 1
                else:
                    set_i.append(str(count)+current_char)
                    count = 1
                    current_char = current_str[x+1]
            current_str = "".join(set_i)
        return current_str

My main idea is to take the first char and check whether the next char is same or not.
If same, count plus one.
If not, add count and current_char into set.
Then reset the count and current_char and repeat.