从两端生成具有相同大小 LIS 的 N 长度排列
给定一个整数N ，任务是生成范围[1, N]中元素的排列，使得LIS从开始的长度等于 LIS 从数组末尾的长度。如果不存在这样的数组，则打印 -1。
例子：
Input: N = 5
Output: [1, 3, 5, 4, 2]
Explanation:
LIS from left side: [1, 3, 5]
LIS from right side: [2, 4, 5]
Length of LIS from left and right side is same that is 3.
Input: N = 7
Output: [1, 3, 4, 7, 6, 5, 2]
Explanation:
LIS from left side: [1, 3, 4, 7]
LIS from right side: [2, 5, 6, 7]
Length of LIS from left and right side is same that is 4.
方法：使用下面的观察来解决这个问题：
- If N = 2, then such an array doesn’t exist.
- Else if N is odd,
- Then start by adding 1 at index 0 and adding 2 at index N-1.
- Now, add N at index N/2. After that, if N>3 then add remaining numbers 3, 4, 5, ... so on indices 1, 2, …, (N/2-1).
- Now add remaining numbers in decreasing order from index N/2+1 to N-2.
- If N is even,
- say 4 then it has only 1 possible solution: {2, 1, 4, 3}.
- After 4, suppose N = 6, then just add 6 at starting and 5 at ending to the previous answer for N = 4 which forms array = [6, 2, 1, 4, 3, 5].
- Similarly, for next even N just add N at 0th index and N-1 at last index..
请按照以下步骤解决问题：
- 如果提到的大小为 2，则返回 -1。
- 否则，根据上述观察添加元素。
- 返回形成的列表并打印列表元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to create array having
// same LIS from both sides
vector equalLIS(int N)
{
// If N = 2 we can't create array
if (N == 2) {
vector al;
al.push_back(-1);
return al;
}
// If N is odd it is possible
// to make an array
else if (N % 2 != 0) {
// Hash Map contains index
// with its value
map lis;
// indices 0, N/2, and N-1
lis.insert({ 0, 1 });
lis.insert({ N - 1, 2 });
int mid = N / 2;
lis.insert({ mid, N });
if (N > 3) {
// Adding numbers 3, 4, ...
// and so on at index from
// 1 to mid-1 respectively
int val = 3;
for (int i = 1; i < mid; i++) {
lis.insert({ i, val });
val++;
}
// Adding remaining integers
// in decreasing order
// from mid+1 to N-2
val = N - 1;
for (int i = mid + 1; i < N - 1; i++) {
lis.insert({ i, val });
val--;
}
}
// Creating Array List
// which will use to form
// required array
vector al;
// Traversing from smallest key
// to largest key in Hash Map
// and adding its value in list
for (int i = 0; i < N; i++) {
al.push_back(lis[i]);
}
// Returning formed array
return al;
}
else {
// Hash Map which contains
// index with its value
map lis;
// Adding value for N=4 in Hash Map
lis.insert({ 0, 2 });
lis.insert({ 1, 1 });
lis.insert({ 2, 4 });
lis.insert({ 3, 3 });
int i = 3;
int idx = 0;
if (N >= 6) {
i++;
idx--;
// Adding new N at starting index
// and N-1 at last index
int val = 5;
while (val <= N) {
lis.insert({ i, val });
lis.insert({ idx, val + 1 });
idx--;
i++;
val += 2;
}
}
// Creating Array List
// which will use to form
// required array
vector al;
// Traversing from minimum key
// to maximum key add
// adding its value in Array List
for (int j = idx + 1; j < i; j++) {
al.push_back(lis[j]);
}
// Returning formed array
return al;
}
}
// Driver Code
int main()
{
int N = 7;
vector lis = equalLIS(N);
for (auto x : lis)
cout << x << " ";
return 0;
}
// This code is contributed by rakeshsahni Java
// Java program for the above approach
import java.util.*;
class GFG {
// Driver Code
public static void main(String[] args)
{
int N = 7;
List lis = equalLIS(N);
for (int x : lis)
System.out.print(x + " ");
}
// Function to create array having
// same LIS from both sides
public static List equalLIS(int N)
{
// If N = 2 we can't create array
if (N == 2) {
List al = new ArrayList<>();
al.add(-1);
return al;
}
// If N is odd it is possible
// to make an array
else if (N % 2 != 0) {
// Hash Map contains index
// with its value
Map lis
= new HashMap<>();
// Adding 1, N, and 2 at
// indices 0, N/2, and N-1
lis.put(0, 1);
lis.put(N - 1, 2);
int mid = N / 2;
lis.put(mid, N);
if (N > 3) {
// Adding numbers 3, 4, ...
// and so on at index from
// 1 to mid-1 respectively
int val = 3;
for (int i = 1; i < mid; i++) {
lis.put(i, val);
val++;
}
// Adding remaining integers
// in decreasing order
// from mid+1 to N-2
val = N - 1;
for (int i = mid + 1;
i < N - 1; i++) {
lis.put(i, val);
val--;
}
}
// Creating Array List
// which will use to form
// required array
List al
= new ArrayList<>();
// Traversing from smallest key
// to largest key in Hash Map
// and adding its value in list
for (int i = 0; i < N; i++) {
al.add(lis.get(i));
}
// Returning formed array
return al;
}
else {
// Hash Map which contains
// index with its value
Map lis
= new HashMap<>();
// Adding value for N=4 in Hash Map
lis.put(0, 2);
lis.put(1, 1);
lis.put(2, 4);
lis.put(3, 3);
int i = 3;
int idx = 0;
if (N >= 6) {
i++;
idx--;
// Adding new N at starting index
// and N-1 at last index
int val = 5;
while (val <= N) {
lis.put(i, val);
lis.put(idx, val + 1);
idx--;
i++;
val += 2;
}
}
// Creating Array List
// which will use to form
// required array
List al
= new ArrayList<>();
// Traversing from minimum key
// to maximum key add
// adding its value in Array List
for (int j = idx + 1; j < i; j++) {
al.add(lis.get(j));
}
// Returning formed array
return al;
}
}
} C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Driver Code
public static void Main(string[] args)
{
int N = 7;
List lis = equalLIS(N);
for (int i = 0; i < lis.Count; i++)
Console.Write(lis[i] + " ");
}
// Function to create array having
// same LIS from both sides
public static List equalLIS(int N)
{
// If N = 2 we can't create array
if (N == 2) {
List al = new List();
al.Add(-1);
return al;
}
// If N is odd it is possible
// to make an array
else if (N % 2 != 0) {
// Hash Map contains index
// with its value
Dictionary lis
= new Dictionary();
// Adding 1, N, and 2 at
// indices 0, N/2, and N-1
lis[0] = 1;
lis[N - 1] = 2;
int mid = N / 2;
lis[mid] = N;
if (N > 3) {
// Adding numbers 3, 4, ...
// and so on at index from
// 1 to mid-1 respectively
int val = 3;
for (int i = 1; i < mid; i++) {
lis[i] = val;
val++;
}
// Adding remaining integers
// in decreasing order
// from mid+1 to N-2
val = N - 1;
for (int i = mid + 1; i < N - 1; i++) {
lis[i] = val;
val--;
}
}
// Creating Array List
// which will use to form
// required array
List al = new List();
// Traversing from smallest key
// to largest key in Hash Map
// and adding its value in list
for (int i = 0; i < N; i++) {
al.Add(lis[i]);
}
// Returning formed array
return al;
}
else {
// Hash Map which contains
// index with its value
Dictionary lis
= new Dictionary();
// Adding value for N=4 in Hash Map
lis[0] = 2;
lis[1] = 1;
lis[2] = 4;
lis[3] = 3;
int i = 3;
int idx = 0;
if (N >= 6) {
i++;
idx--;
// Adding new N at starting index
// and N-1 at last index
int val = 5;
while (val <= N) {
lis[i] = val;
lis[idx] = val + 1;
idx -= 1;
i += 1;
val += 2;
}
}
// Creating Array List
// which will use to form
// required array
List al = new List();
// Traversing from minimum key
// to maximum key add
// adding its value in Array List
for (int j = idx + 1; j < i; j++) {
al.Add(lis[j]);
}
// Returning formed array
return al;
}
}
}
// This code is contributed by ukasp. 输出
1 3 4 7 6 5 2 时间复杂度： 在）
辅助空间： O(N)