编写一个算法来返回一个数组，使来自1..n的每个数字k正好出现两次，并且距离其副本是k个距离 [英] Write an algorithm to return an array such that every number k from 1..n is occurring exactly twice and is k distance apart from its replica

问题描述

对于给定的整数n> = 3，返回一个大小为2n的数组，使得从1到n的每个数字k都是发生恰好两次，每个数字，重复的距离等于该数字。

功能签名：

  int * buildArray（int n）
  

对于n = 3：

  3，1，2，1，3，2 
  

号码 2 ：第1位3和第2位6，所以距离6 - 3 - 1 = 2.

号码 3 ：第一个 3 在位置1和第二个 3 在位置5，所以距离5 - 1 - 1 = 3。

对于n = 4：

  4，1，3，1，2，4，3，2 
  

解决方案

这是一个Langford的问题/序列。

有一个关于SO已经实现了同样的问题。
Langford序列实现Haskell或C

This question was asked in an interview.

For a given integer n >= 3 return an array of size 2n such that every number k from 1 to n is occurring exactly twice and every number and its repetition is separated by a distance equal to the number.

Function signature:

int* buildArray(int n)

For example, for n = 3:

3, 1, 2, 1, 3, 2

Number 2: 1st position 3 and 2nd position 6, so distance 6 - 3 - 1 = 2.
Number 3: First 3 at position 1 and 2nd 3 at position 5, so distance 5 - 1 - 1 = 3.

For n = 4:

4, 1, 3, 1, 2, 4, 3, 2

解决方案

It's a Langford's problem/sequence.

There is a topic about the same problem on SO with implementation already. Langford sequence implementation Haskell or C

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