给定一个长度为N且具有唯一元素的数组arr[] ，任务是找到可以由数组两端的元素组成的最长递增子序列的长度。
例子：
Input: arr[] = {3, 5, 1, 4, 2}
Output: 4
Explanation:
The longest sequence is: {2, 3, 4, 5}
Pick 2, Sequence is {2}, Array is {3, 5, 1, 4}
Pick 3, Sequence is {2, 3}, Array is {5, 1, 4}
Pick 4, Sequence is {2, 3, 4}, Array is {5, 1}
Pick 5, Sequence is {2, 3, 4, 5}, Array is {1}
Input: arr[] = {3, 1, 5, 2, 4}
Output: 2
The longest sequence is {3, 4}
方法：这个问题可以通过两个指针方法来解决。在数组的第一个和最后一个索引处设置两个指针。选择当前指向的两个值中的最小值并检查它是否大于先前选择的值。如果是，则更新指针并增加 LIS 的长度并重复该过程。否则，打印获得的 LIS 的长度。
下面是上述方法的实现：
C++
// C++ Program to print the
// longest increasing
// subsequence from the
// boundary elements of an array
#include
using namespace std;
// Function to return the length of
// Longest Increasing subsequence
int longestSequence(int n, int arr[])
{
// Set pointers at
// both ends
int l = 0, r = n - 1;
// Stores the recent
// value added to the
// subsequence
int prev = INT_MIN;
// Stores the length of
// the subsequence
int ans = 0;
while (l <= r) {
// Check if both elements
// can be added to the
// subsequence
if (arr[l] > prev
&& arr[r] > prev) {
if (arr[l] < arr[r]) {
ans += 1;
prev = arr[l];
l += 1;
}
else {
ans += 1;
prev = arr[r];
r -= 1;
}
}
// Check if the element
// on the left can be
// added to the
// subsequence only
else if (arr[l] > prev) {
ans += 1;
prev = arr[l];
l += 1;
}
// Check if the element
// on the right can be
// added to the
// subsequence only
else if (arr[r] > prev) {
ans += 1;
prev = arr[r];
r -= 1;
}
// If none of the values
// can be added to the
// subsequence
else {
break;
}
}
return ans;
}
// Driver Code
int main()
{
int arr[] = { 3, 5, 1, 4, 2 };
// Length of array
int n = sizeof(arr)
/ sizeof(arr[0]);
cout << longestSequence(n, arr);
return 0;
} Java
// Java program to print the longest
// increasing subsequence from the
// boundary elements of an array
import java.util.*;
class GFG{
// Function to return the length of
// Longest Increasing subsequence
static int longestSequence(int n, int arr[])
{
// Set pointers at
// both ends
int l = 0, r = n - 1;
// Stores the recent
// value added to the
// subsequence
int prev = Integer.MIN_VALUE;
// Stores the length of
// the subsequence
int ans = 0;
while (l <= r)
{
// Check if both elements
// can be added to the
// subsequence
if (arr[l] > prev &&
arr[r] > prev)
{
if (arr[l] < arr[r])
{
ans += 1;
prev = arr[l];
l += 1;
}
else
{
ans += 1;
prev = arr[r];
r -= 1;
}
}
// Check if the element on the
// left can be added to the
// subsequence only
else if (arr[l] > prev)
{
ans += 1;
prev = arr[l];
l += 1;
}
// Check if the element on the
// right can be added to the
// subsequence only
else if (arr[r] > prev)
{
ans += 1;
prev = arr[r];
r -= 1;
}
// If none of the values
// can be added to the
// subsequence
else
{
break;
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 3, 5, 1, 4, 2 };
// Length of array
int n = arr.length;
System.out.print(longestSequence(n, arr));
}
}
// This code is contributed by Amit KatiyarPython3
# Python3 program to print the
# longest increasing subsequence
# from the boundary elements
# of an array
import sys
# Function to return the length of
# Longest Increasing subsequence
def longestSequence(n, arr):
# Set pointers at
# both ends
l = 0
r = n - 1
# Stores the recent value
# added to the subsequence
prev = -sys.maxsize - 1
# Stores the length of
# the subsequence
ans = 0
while (l <= r):
# Check if both elements can be
# added to the subsequence
if (arr[l] > prev and
arr[r] > prev):
if (arr[l] < arr[r]):
ans += 1
prev = arr[l]
l += 1
else:
ans += 1
prev = arr[r]
r -= 1
# Check if the element
# on the left can be
# added to the
# subsequence only
elif (arr[l] > prev):
ans += 1
prev = arr[l]
l += 1
# Check if the element
# on the right can be
# added to the
# subsequence only
elif (arr[r] > prev):
ans += 1
prev = arr[r]
r -= 1
# If none of the values
# can be added to the
# subsequence
else:
break
return ans
# Driver code
arr = [ 3, 5, 1, 4, 2 ]
# Length of array
n = len(arr)
print(longestSequence(n, arr))
# This code is contributed by sanjoy_62C#
// C# program to print the longest
// increasing subsequence from the
// boundary elements of an array
using System;
class GFG{
// Function to return the length of
// longest Increasing subsequence
static int longestSequence(int n, int []arr)
{
// Set pointers at
// both ends
int l = 0, r = n - 1;
// Stores the recent value
// added to the subsequence
int prev = int.MinValue;
// Stores the length of
// the subsequence
int ans = 0;
while (l <= r)
{
// Check if both elements
// can be added to the
// subsequence
if (arr[l] > prev &&
arr[r] > prev)
{
if (arr[l] < arr[r])
{
ans += 1;
prev = arr[l];
l += 1;
}
else
{
ans += 1;
prev = arr[r];
r -= 1;
}
}
// Check if the element on the
// left can be added to the
// subsequence only
else if (arr[l] > prev)
{
ans += 1;
prev = arr[l];
l += 1;
}
// Check if the element on the
// right can be added to the
// subsequence only
else if (arr[r] > prev)
{
ans += 1;
prev = arr[r];
r -= 1;
}
// If none of the values
// can be added to the
// subsequence
else
{
break;
}
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 3, 5, 1, 4, 2 };
// Length of array
int n = arr.Length;
Console.Write(longestSequence(n, arr));
}
}
// This code is contributed by Amit KatiyarJavascript
输出：
4时间复杂度： O(N)