给定一个包含两个已合并序列的合并序列，其中一个严格增加，另一个严格减少。在不更改顺序的元素之间插入顺序递增的元素。
Sequences [1, 3, 4] and [10, 4, 2] can produce the following resulting sequences:
[10, 1, 3, 4, 2, 4], [1, 3, 4, 10, 4, 2].
The following sequence cannot be the result of these insertions:
[1, 10, 4, 4, 3, 2] because the order of elements in the increasing sequence was changed.
给定一个合并的序列，任务是找到任何两个合适的初始序列，其中一个应严格增加，而另一个应严格减少。
注意：空序列和由一个元素组成的序列可以视为递增或递减。
例子：
Input: arr[] = {5, 1, 3, 6, 8, 2, 9, 0, 10}
Output: [1, 3, 6, 8, 9, 10] [5, 2, 0]
Input: arr[] = {1, 2, 4, 0, 2}
Output: -1
No such sequences possible.
方法1：我们可以修改最长递增序列)并解决所需的问题。这将花费O(nlogn)时间。
方法2：我们也可以只在单个遍历中解决此问题。这里使用的想法是维护两个排序的数组。
对于新元素x ，
- 如果只能将其附加到数组之一，则附加它。
- 如果不能将两者都附加，则答案为-1 。
- 如果它可以被附加到两个再检查下一个元素y，如果y> X，则追加x到增加一个附加否则x到减小一个。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print strictly increasing and
// strictly decreasing sequence if possible
void Find_Sequence(int arr[], int n)
{
// Arrays to store strictly increasing and
// decreasing sequence
vector inc_arr, dec_arr;
// Initializing last element of both sequence
int flag = 0;
long inc = -1, dec = 1e7;
// Iterating through the array
for (int i = 0; i < n; i++)
{
// If current element can be appended
// to both the sequences
if (inc < arr[i] && arr[i] < dec)
{
// If next element is greater than
// the current element
// Then append it to the strictly
// increasing array
if (arr[i] < arr[i + 1])
{
inc = arr[i];
inc_arr.emplace_back(arr[i]);
}
// Otherwise append it to the
// strictly decreasing array
else
{
dec = arr[i];
dec_arr.emplace_back(arr[i]);
}
}
// If current element can be appended
// to the increasing sequence only
else if (inc < arr[i])
{
inc = arr[i];
inc_arr.emplace_back(arr[i]);
}
// If current element can be appended
// to the decreasing sequence only
else if (dec > arr[i])
{
dec = arr[i];
dec_arr.emplace_back(arr[i]);
}
// Else we can not make such sequences
// from the given array
else
{
cout << -1 << endl;
flag = 1;
break;
}
}
// Print the required sequences
if (!flag)
{
for (auto i = inc_arr.begin();
i != inc_arr.end(); i++)
cout << *i << " ";
cout << endl;
for (auto i = dec_arr.begin();
i != dec_arr.end(); i++)
cout << *i << " ";
cout << endl;
}
}
// Driver code
int main()
{
int arr[] = { 5, 1, 3, 6, 8, 2, 9, 0, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
Find_Sequence(arr, n);
}
// This code is contributed by sanjeev2552 Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to print strictly increasing and
// strictly decreasing sequence if possible
static void Find_Sequence(int[] arr, int n)
{
// Arrays to store strictly increasing and
// decreasing sequence
Vector inc_arr = new Vector<>(),
dec_arr = new Vector<>();
// Initializing last element of both sequence
int flag = 0;
long inc = -1, dec = (long) 1e7;
// Iterating through the array
for (int i = 0; i < n; i++)
{
// If current element can be appended
// to both the sequences
if (inc < arr[i] && arr[i] < dec)
{
// If next element is greater than
// the current element
// Then append it to the strictly
// increasing array
if (arr[i] < arr[i + 1])
{
inc = arr[i];
inc_arr.add(arr[i]);
}
// Otherwise append it to the
// strictly decreasing array
else
{
dec = arr[i];
dec_arr.add(arr[i]);
}
}
// If current element can be appended
// to the increasing sequence only
else if (inc < arr[i])
{
inc = arr[i];
inc_arr.add(arr[i]);
}
// If current element can be appended
// to the decreasing sequence only
else if (dec > arr[i])
{
dec = arr[i];
dec_arr.add(arr[i]);
}
// Else we can not make such sequences
// from the given array
else
{
System.out.println(-1);
flag = 1;
break;
}
}
// Print the required sequences
if (flag == 0)
{
for (int i : inc_arr)
System.out.print(i + " ");
System.out.println();
for (int i : dec_arr)
System.out.print(i + " ");
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 5, 1, 3, 6, 8, 2, 9, 0, 10 };
int n = arr.length;
Find_Sequence(arr, n);
}
}
// This code is contributed by
// sanjeev2552 Python3
# Python3 implementation of the approach
# Function to print strictly increasing and
# strictly decreasing sequence if possible
def Find_Sequence(array, n):
# Arrays to store strictly increasing and
# decreasing sequence
inc_arr, dec_arr =[], []
# Initializing last element of both sequence
inc, dec = -1, 1e7
# Iterating through the array
for i in range(n):
# If current element can be appended
# to both the sequences
if inc < array[i] < dec:
# If next element is greater than
# the current element
# Then append it to the strictly
# increasing array
if array[i] < array[i + 1]:
inc = array[i]
inc_arr.append(array[i])
# Otherwise append it to the
# strictly decreasing array
else:
dec = array[i]
dec_arr.append(array[i])
# If current element can be appended
# to the increasing sequence only
elif inc < array[i]:
inc = array[i]
inc_arr.append(array[i])
# If current element can be appended
# to the decreasing sequence only
elif dec > array[i]:
dec = array[i]
dec_arr.append(array[i])
# Else we can not make such sequences
# from the given array
else:
print('-1')
break
# Print the required sequences
else:
print(inc_arr, dec_arr)
# Driver code
arr = [5, 1, 3, 6, 8, 2, 9, 0, 10]
n = len(arr)
Find_Sequence(arr, n)C#
// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// Function to print strictly increasing and
// strictly decreasing sequence if possible
static void Find_Sequence(int[] arr, int n)
{
// Arrays to store strictly increasing and
// decreasing sequence
ArrayList inc_arr = new ArrayList();
ArrayList dec_arr = new ArrayList();
// Initializing last element of both sequence
int flag = 0;
long inc = -1, dec = (long)1e7;
// Iterating through the array
for(int i = 0; i < n; i++)
{
// If current element can be appended
// to both the sequences
if (inc < arr[i] && arr[i] < dec)
{
// If next element is greater than
// the current element
// Then append it to the strictly
// increasing array
if (arr[i] < arr[i + 1])
{
inc = arr[i];
inc_arr.Add(arr[i]);
}
// Otherwise append it to the
// strictly decreasing array
else
{
dec = arr[i];
dec_arr.Add(arr[i]);
}
}
// If current element can be appended
// to the increasing sequence only
else if (inc < arr[i])
{
inc = arr[i];
inc_arr.Add(arr[i]);
}
// If current element can be appended
// to the decreasing sequence only
else if (dec > arr[i])
{
dec = arr[i];
dec_arr.Add(arr[i]);
}
// Else we can not make such sequences
// from the given array
else
{
Console.Write(-1);
flag = 1;
break;
}
}
// Print the required sequences
if (flag == 0)
{
foreach(int i in inc_arr)
Console.Write(i + " ");
Console.Write('\n');
foreach(int i in dec_arr)
Console.Write(i + " ");
Console.Write('\n');
}
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 5, 1, 3, 6, 8,
2, 9, 0, 10 };
int n = arr.Length;
Find_Sequence(arr, n);
}
}
// This code is contributed by rutvik_56PHP
$arr[$i])
{
$dec = $arr[$i];
array_push($dec_arr, $arr[$i]);
}
// Else we can not make such sequences
// from the given array
else
{
echo '-1';
break;
}
}
// Print the required sequences
print_r($inc_arr);
print_r($dec_arr);
}
// Driver code
$arr = array(5, 1, 3, 6, 8, 2, 9, 0, 10);
$n = count($arr);
Find_Sequence($arr, $n);
// This code is contributed by Ryuga
?>输出：
[1, 3, 6, 8, 9, 10] [5, 2, 0]