{ "data": { "question": { "questionId": "3783", "questionFrontendId": "3470", "categoryTitle": "Algorithms", "boundTopicId": 3590023, "title": "Permutations IV", "titleSlug": "permutations-iv", "content": "
Given two integers, n and k, an alternating permutation is a permutation of the first n positive integers such that no two adjacent elements are both odd or both even.
Return the k-th alternating permutation sorted in lexicographical order. If there are fewer than k valid alternating permutations, return an empty list.
\n
Example 1:
\n\nInput: n = 4, k = 6
\n\nOutput: [3,4,1,2]
\n\nExplanation:
\n\nThe lexicographically-sorted alternating permutations of [1, 2, 3, 4] are:
[1, 2, 3, 4][1, 4, 3, 2][2, 1, 4, 3][2, 3, 4, 1][3, 2, 1, 4][3, 4, 1, 2] ← 6th permutation[4, 1, 2, 3][4, 3, 2, 1]Since k = 6, we return [3, 4, 1, 2].
Example 2:
\n\nInput: n = 3, k = 2
\n\nOutput: [3,2,1]
\n\nExplanation:
\n\nThe lexicographically-sorted alternating permutations of [1, 2, 3] are:
[1, 2, 3][3, 2, 1] ← 2nd permutationSince k = 2, we return [3, 2, 1].
Example 3:
\n\nInput: n = 2, k = 3
\n\nOutput: []
\n\nExplanation:
\n\nThe lexicographically-sorted alternating permutations of [1, 2] are:
[1, 2][2, 1]There are only 2 alternating permutations, but k = 3, which is out of range. Thus, we return an empty list [].
\n
Constraints:
\n\n1 <= n <= 1001 <= k <= 1015给你两个整数 n 和 k,一个 交替排列 是前 n 个正整数的排列,且任意相邻 两个 元素不都为奇数或都为偶数。
返回第 k 个 交替排列 ,并按 字典序 排序。如果有效的 交替排列 少于 k 个,则返回一个空列表。
\n\n
示例 1
\n\n输入:n = 4, k = 6
\n\n输出:[3,4,1,2]
\n\n解释:
\n\n[1, 2, 3, 4] 的交替排列按字典序排序后为:
[1, 2, 3, 4][1, 4, 3, 2][2, 1, 4, 3][2, 3, 4, 1][3, 2, 1, 4][3, 4, 1, 2] ← 第 6 个排列[4, 1, 2, 3][4, 3, 2, 1]由于 k = 6,我们返回 [3, 4, 1, 2]。
示例 2
\n\n输入:n = 3, k = 2
\n\n输出:[3,2,1]
\n\n解释:
\n\n[1, 2, 3] 的交替排列按字典序排序后为:
[1, 2, 3][3, 2, 1] ← 第 2 个排列由于 k = 2,我们返回 [3, 2, 1]。
示例 3
\n\n输入:n = 2, k = 3
\n\n输出:[]
\n\n解释:
\n\n[1, 2] 的交替排列按字典序排序后为:
[1, 2][2, 1]只有 2 个交替排列,但 k = 3 超出了范围。因此,我们返回一个空列表 []。
\n\n
提示:
\n\n1 <= n <= 1001 <= k <= 1015n is odd, the first number must be odd.",
"If n is even, the first number can be either odd or even.",
"From smallest to largest, place each number and subtract the number of permutations from k.",
"The number of permutations can be calculated using factorials."
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