给定大小为N的数组A [] ，任务是通过将给定数组两端的元素附加到数组末尾，找到可以从给定数组生成的最长非递减数组的最大长度。结果数组并一一删除附加元素。
例子：
Input: A[] = {5, 6, 8, 7}
Output:: 4
Explanation:
If resArr[] = { } is a newly generated array then
Inserting A[0] into resArr[] modifies A[] = {6, 8, 7} and resArr[] = {5}
Inserting A[0] into resArr[] modifies A[] = {8, 7} and resArr[] = {5, 6}
Inserting A[1] into resArr[] modifies A[] = {8} and resArr[] = {5, 6, 7}
Inserting A[0] into resArr[] modifies A[] = {} and resArr[] = {5, 6, 7, 8}
Therefore, the length of the longest non-decreasing array that can be generated from the given array is 4.
Input: {1, 1, 3, 5, 4, 3, 6, 2, 1}
Output: 7
方法：可以使用两个指针方法解决此问题。请按照以下步骤解决问题：
- 初始化一个变量，例如res = 0，以存储可以从给定数组生成的非递减数组的最大长度。
- 初始化两个变量，例如start = 0和end = N – 1，以存储开始和结束指针的索引。
- 遍历数组并检查以下条件：
- 如果A [start] <= A [end]，则检查A [start]的值是否大于或等于新数组中先前插入的元素。如果确定为true，则将start和res的值增加1 。
- 如果A [开始]> = A [结束]然后检查是否A [结束]的值是大于或等于所述新的数组或没有在先前插入的元件。如果确定为true，则将end的值减1，并将res的值减1 。
- 最后，打印res的值。
下面是上述方法的实现：
C++
//C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the length of the longest
// non-decreasing array that can be generated
int findLongestNonDecreasing(int A[], int N)
{
// Stores the length of the longest
// non-decreasing array that can be
// generated from the array
int res = 0;
// Stores index of
// start pointer
int start = 0;
// Stores index of
// end pointer
int end = N - 1;
// Stores previously inserted
// element into the new array
int prev = -1;
// Traverse the array
while (start <= end) {
// If A[start] is less than
// or equal to A[end]
if (A[start] <= A[end]) {
// If no element inserted into
// the newly generated array
if (prev == -1) {
// Update prev
prev = A[start];
// Update res
res++;
// Update start
start++;
}
else {
// If A[start] is greater
// than or equal to prev
if (A[start] >= prev) {
// Update res
res++;
// Update prev
prev = A[start];
// Update start
start++;
}
// If A[end] is greater
// than or equal to prev
else if (A[end] >= prev) {
// Update res
res++;
// Update prev
prev = A[end];
// Update end
end--;
}
else {
break;
}
}
}
// If A[end] is
// greater than A[start]
else{
// If no element inserted into
// the newly generated array
if (prev == -1) {
// Update prev
prev = A[end];
// Update res
res++;
// Update end
end--;
}
else {
// If A[end] is greater
// than or equal to prev
if (A[end] >= prev) {
// Update res
res++;
//Update prev
prev = A[end];
// Update end
end--;
}
// If A[start] is greater
// than or equal to prev
else if (A[start] >= prev) {
// Update res
res++;
//Update prev
prev = A[start];
// Update start
start++;
}
else {
break;
}
}
}
}
return res;
}
//Driver Code
int main()
{
int A[]={ 1, 1, 3, 5, 4, 3, 6, 2, 1 };
int N = sizeof(A)/sizeof(A[0]);
//Function call
cout<< findLongestNonDecreasing(A, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to find the length of the longest
// non-decreasing array that can be generated
static int findLongestNonDecreasing(int A[],
int N)
{
// Stores the length of the longest
// non-decreasing array that can be
// generated from the array
int res = 0;
// Stores index of
// start pointer
int start = 0;
// Stores index of
// end pointer
int end = N - 1;
// Stores previously inserted
// element into the new array
int prev = -1;
// Traverse the array
while (start <= end)
{
// If A[start] is less than
// or equal to A[end]
if (A[start] <= A[end])
{
// If no element inserted into
// the newly generated array
if (prev == -1)
{
// Update prev
prev = A[start];
// Update res
res++;
// Update start
start++;
}
else
{
// If A[start] is greater
// than or equal to prev
if (A[start] >= prev)
{
// Update res
res++;
// Update prev
prev = A[start];
// Update start
start++;
}
// If A[end] is greater
// than or equal to prev
else if (A[end] >= prev)
{
// Update res
res++;
// Update prev
prev = A[end];
// Update end
end--;
}
else
{
break;
}
}
}
// If A[end] is
// greater than A[start]
else
{
// If no element inserted into
// the newly generated array
if (prev == -1)
{
// Update prev
prev = A[end];
// Update res
res++;
// Update end
end--;
}
else
{
// If A[end] is greater
// than or equal to prev
if (A[end] >= prev)
{
// Update res
res++;
//Update prev
prev = A[end];
// Update end
end--;
}
// If A[start] is greater
// than or equal to prev
else if (A[start] >= prev)
{
// Update res
res++;
//Update prev
prev = A[start];
// Update start
start++;
}
else
{
break;
}
}
}
}
return res;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 1, 1, 3, 5, 4, 3, 6, 2, 1 };
int N = A.length;
// Function call
System.out.print(findLongestNonDecreasing(A, N));
}
}
// This code is contributed by susmitakundugoaldangaPython3
# Python3 program to implement
# the above approach
# Function to find the length of
# the longest non-decreasing array
# that can be generated
def findLongestNonDecreasing(A, N):
# Stores the length of the longest
# non-decreasing array that can be
# generated from the array
res = 0;
# Stores index of
# start pointer
start = 0;
# Stores index of
# end pointer
end = N - 1;
# Stores previously inserted
# element into the new array
prev = -1;
# Traverse the array
while (start <= end):
# If A[start] is less than
# or equal to A[end]
if (A[start] <= A[end]):
# If no element inserted into
# the newly generated array
if (prev == -1):
# Update prev
prev = A[start];
# Update res
res += 1
# Update start
start += 1
else:
# If A[start] is greater
# than or equal to prev
if (A[start] >= prev):
# Update res
res += 1
# Update prev
prev = A[start];
# Update start
start += 1
# If A[end] is greater
# than or equal to prev
elif (A[end] >= prev):
# Update res
res += 1
# Update prev
prev = A[end];
# Update end
end -= 1
else:
break;
# If A[end] is
# greater than A[start]
else:
# If no element inserted into
# the newly generated array
if (prev == -1):
# Update prev
prev = A[end];
# Update res
res += 1
# Update end
end -= 1
else:
# If A[end] is greater
# than or equal to prev
if (A[end] >= prev):
# Update res
res += 1
# Update prev
prev = A[end];
# Update end
end -= 1
# If A[start] is greater
# than or equal to prev
elif (A[start] >= prev):
# Update res
res += 1
# Update prev
prev = A[start];
# Update start
start += 1
else :
break;
return res
# Driver Code
if __name__ == "__main__":
A = [1, 1, 3, 5,
4, 3, 6, 2, 1]
N = len(A)
# Function call
print (findLongestNonDecreasing(A, N));
# This code is contributed by ChitranayalC#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to find the length of the longest
// non-decreasing array that can be generated
static int findLongestNonDecreasing(int[] A,
int N)
{
// Stores the length of the longest
// non-decreasing array that can be
// generated from the array
int res = 0;
// Stores index of
// start pointer
int start = 0;
// Stores index of
// end pointer
int end = N - 1;
// Stores previously inserted
// element into the new array
int prev = -1;
// Traverse the array
while (start <= end)
{
// If A[start] is less than
// or equal to A[end]
if (A[start] <= A[end])
{
// If no element inserted into
// the newly generated array
if (prev == -1)
{
// Update prev
prev = A[start];
// Update res
res++;
// Update start
start++;
}
else
{
// If A[start] is greater
// than or equal to prev
if (A[start] >= prev)
{
// Update res
res++;
// Update prev
prev = A[start];
// Update start
start++;
}
// If A[end] is greater
// than or equal to prev
else if (A[end] >= prev)
{
// Update res
res++;
// Update prev
prev = A[end];
// Update end
end--;
}
else
{
break;
}
}
}
// If A[end] is
// greater than A[start]
else
{
// If no element inserted into
// the newly generated array
if (prev == -1)
{
// Update prev
prev = A[end];
// Update res
res++;
// Update end
end--;
}
else
{
// If A[end] is greater
// than or equal to prev
if (A[end] >= prev)
{
// Update res
res++;
//Update prev
prev = A[end];
// Update end
end--;
}
// If A[start] is greater
// than or equal to prev
else if (A[start] >= prev)
{
// Update res
res++;
//Update prev
prev = A[start];
// Update start
start++;
}
else
{
break;
}
}
}
}
return res;
}
// Driver Code
public static void Main()
{
int[] A = { 1, 1, 3, 5, 4, 3, 6, 2, 1 };
int N = A.Length;
// Function call
Console.Write(findLongestNonDecreasing(A, N));
}
}
// This code is contributed by sanjoy_62输出：
7
时间复杂度： O(N)
空间复杂度： O(1)