Find palindrome with fixed length

class Solution:
    def kthPalindrome(self, queries: List[int], intLength: int) -> List[int]:
        l = (intLength + 1) // 2   # 可以唯一确定回文数的前半部分的长度
        start = 10 ** (l - 1) - 1   # start + k 即为第 k 个 l 位无前导零整数
        limit = 10 ** l - 1   # l 位无前导零整数的上界
        res = []
        # 将前半部分恢复为对应的回文数
        def recover(num: int) -> int:
            if intLength % 2 == 0:
                return int(str(num) + str(num)[::-1])
            else:
                return int(str(num)[:-1] + str(num)[::-1])

        # 依次处理询问
        for query in queries:
            if start + query > limit:
                # 不存在
                res.append(-1)
                continue
            res.append(recover(start + query))
        return res

Find Palindrome With Fixed Length

Difficulty: Medium

Given an integer array queries and a positive integer intLength, return an array answer where answer[i] is either the queries[i]th smallest positive palindrome of length intLength or -1 if no such palindrome exists.

A palindrome is a number that reads the same backwards and forwards. Palindromes cannot have leading zeros.

 

Example 1:

Input: queries = [1,2,3,4,5,90], intLength = 3
Output: [101,111,121,131,141,999]
Explanation:
The first few palindromes of length 3 are:
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ...
The 90th palindrome of length 3 is 999.

Example 2:

Input: queries = [2,4,6], intLength = 4
Output: [1111,1331,1551]
Explanation:
The first six palindromes of length 4 are:
1001, 1111, 1221, 1331, 1441, and 1551.

 

Constraints:

  • 1 <= queries.length <= 5 * 104
  • 1 <= queries[i] <= 109
  • 1 <= intLength <= 15