安排前N个自然数，使得相邻元素的绝对差大于1

我们有前 N 个自然数。我们的任务是获得它们的一种排列，其中每两个连续元素之间的绝对差 > 1。如果不存在这样的排列，则返回 -1。

方法很简单。我们将使用贪心方法。我们将所有奇数按升序或降序排列，然后将所有偶数按降序或升序排列

算法

arrangeN(n)

Begin
   if N is 1, then return 1
   if N is 2 or 3, then return -1 as no such permutation is not present
   even_max and odd_max is set as max even and odd number less or equal to n
   arrange all odd numbers in descending order
   arrange all even numbers in descending order
End

Example

的中文翻译为：

示例

#include <iostream>
using namespace std;
void arrangeN(int N) {
   if (N == 1) { //if N is 1, only that will be placed
      cout << "1";
      return;
   }
   if (N == 2 || N == 3) { //for N = 2 and 3, no such permutation is available
      cout << "-1";
      return;
   }
   int even_max = -1, odd_max = -1;
   //find max even and odd which are less than or equal to N
   if (N % 2 == 0) {
      even_max = N;
      odd_max = N - 1;
   } else {
      odd_max = N;
      even_max = N - 1;
   }
   while (odd_max >= 1) { //print all odd numbers in decreasing order
      cout << odd_max << " ";
      odd_max -= 2;
   }
   while (even_max >= 2) { //print all even numbers in decreasing order
      cout << even_max << " ";
      even_max -= 2;
   }
}
int main() {
   int N = 8;
   arrangeN(N);
}

输出

7 5 3 1 8 6 4 2

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