{ "data": { "question": { "questionId": "38", "questionFrontendId": "38", "categoryTitle": "Algorithms", "boundTopicId": 978, "title": "Count and Say", "titleSlug": "count-and-say", "content": "
The count-and-say sequence is a sequence of digit strings defined by the recursive formula:
\n\ncountAndSay(1) = "1"
countAndSay(n)
is the run-length encoding of countAndSay(n - 1)
.Run-length encoding (RLE) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string "3322251"
we replace "33"
with "23"
, replace "222"
with "32"
, replace "5"
with "15"
and replace "1"
with "11"
. Thus the compressed string becomes "23321511"
.
Given a positive integer n
, return the nth
element of the count-and-say sequence.
\n
Example 1:
\n\nInput: n = 4
\n\nOutput: "1211"
\n\nExplanation:
\n\n\ncountAndSay(1) = "1"\ncountAndSay(2) = RLE of "1" = "11"\ncountAndSay(3) = RLE of "11" = "21"\ncountAndSay(4) = RLE of "21" = "1211"\n\n
Example 2:
\n\nInput: n = 1
\n\nOutput: "1"
\n\nExplanation:
\n\nThis is the base case.
\n\n
Constraints:
\n\n1 <= n <= 30
\nFollow up: Could you solve it iteratively?", "translatedTitle": "外观数列", "translatedContent": "
「外观数列」是一个数位字符串序列,由递归公式定义:
\n\ncountAndSay(1) = \"1\"
countAndSay(n)
是 countAndSay(n-1)
的行程长度编码。\n\n
行程长度编码(RLE)是一种字符串压缩方法,其工作原理是通过将连续相同字符(重复两次或更多次)替换为字符重复次数(运行长度)和字符的串联。例如,要压缩字符串 \"3322251\"
,我们将 \"33\"
用 \"23\"
替换,将 \"222\"
用 \"32\"
替换,将 \"5\"
用 \"15\"
替换并将 \"1\"
用 \"11\"
替换。因此压缩后字符串变为 \"23321511\"
。
给定一个整数 n
,返回 外观数列 的第 n
个元素。
示例 1:
\n\n输入:n = 4
\n\n输出:\"1211\"
\n\n解释:
\n\ncountAndSay(1) = \"1\"
\n\ncountAndSay(2) = \"1\" 的行程长度编码 = \"11\"
\n\ncountAndSay(3) = \"11\" 的行程长度编码 = \"21\"
\n\ncountAndSay(4) = \"21\" 的行程长度编码 = \"1211\"
\n示例 2:
\n\n输入:n = 1
\n\n输出:\"1\"
\n\n解释:
\n\n这是基本情况。
\n\n\n
提示:
\n\n1 <= n <= 30
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