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
#includeusing 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
