最长共同递增子序列(LCS + LIS)

先决条件：LCS，LIS
给定两个数组，找到最长的公共递增子序列[LCIS]的长度并打印其中一个序列(可能存在多个序列)
假设我们考虑两个数组–
arr1 [] = {3，4，9，1}和
arr2 [] = {5，3，8，9，10，2，1}
我们的答案将是{3，9}，因为这是最长的公共子序列，并且还在增加。

这个想法也是在这里使用动态编程。我们存储以arr2 []的每个索引结尾的最长的公共递增子序列。我们创建一个辅助数组table []，使得table [j]存储以arr2 [j]结尾的LCIS的长度。最后，我们从该表中返回最大值。为了填充该表中的值，我们遍历了arr1 []的所有元素，对于每个元素arr1 [i]，我们遍历了arr2 []的所有元素。如果找到匹配项，则使用当前LCIS的长度更新表[j]。为了保持当前的LCIS，我们将继续检查有效的table [j]值。
下面是查找LCIS长度的程序。

C++

// A C++ Program to find length of the Longest Common
// Increasing Subsequence (LCIS)
#include
using namespace std;
 
// Returns the length and the LCIS of two
// arrays arr1[0..n-1] and arr2[0..m-1]
int LCIS(int arr1[], int n, int arr2[], int m)
{
    // table[j] is going to store length of LCIS
    // ending with arr2[j]. We initialize it as 0,
    int table[m];
    for (int j=0; j table[j])
                    table[j] = current + 1;
 
            /* Now seek for previous smaller common
               element for current element of arr1 */
            if (arr1[i] > arr2[j])
                if (table[j] > current)
                    current = table[j];
        }
    }
 
    // The maximum value in table[] is out result
    int result = 0;
    for (int i=0; i result)
           result = table[i];
 
    return result;
}
 
/* Driver program to test above function */
int main()
{
    int arr1[] = {3, 4, 9, 1};
    int arr2[] = {5, 3, 8, 9, 10, 2, 1};
 
    int n = sizeof(arr1)/sizeof(arr1[0]);
    int m = sizeof(arr2)/sizeof(arr2[0]);
 
    cout << "Length of LCIS is "
         << LCIS(arr1, n, arr2, m);
    return (0);
}

Java

// A Java Program to find length of the Longest
// Common Increasing Subsequence (LCIS)
import java.io.*;
 
class GFG {
 
    // Returns the length and the LCIS of two
    // arrays arr1[0..n-1] and arr2[0..m-1]
    static int LCIS(int arr1[], int n, int arr2[],
                                         int m)
    {
        // table[j] is going to store length of
        // LCIS ending with arr2[j]. We initialize
        // it as 0,
        int table[] = new int[m];
        for (int j = 0; j < m; j++)
            table[j] = 0;
 
        // Traverse all elements of arr1[]
        for (int i = 0; i < n; i++)
        {
            // Initialize current length of LCIS
            int current = 0;
 
            // For each element of arr1[], trvarse
            // all elements of arr2[].
            for (int j = 0; j < m; j++)
            {
                // If both the array have same
                // elements. Note that we don't
                // break the loop here.
                if (arr1[i] == arr2[j])
                    if (current + 1 > table[j])
                        table[j] = current + 1;
 
                /* Now seek for previous smaller
                common element for current
                element of arr1 */
                if (arr1[i] > arr2[j])
                    if (table[j] > current)
                        current = table[j];
            }
        }
 
        // The maximum value in table[] is out
        // result
        int result = 0;
        for (int i=0; i result)
            result = table[i];
 
        return result;
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr1[] = {3, 4, 9, 1};
        int arr2[] = {5, 3, 8, 9, 10, 2, 1};
 
        int n = arr1.length;
        int m = arr2.length;
 
    System.out.println("Length of LCIS is " +
                       LCIS(arr1, n, arr2, m));
    }
}
// This code is contributed by Prerna Saini

Python 3

# Python 3 Program to find length of the
# Longest Common Increasing Subsequence (LCIS)
 
# Returns the length and the LCIS of two
# arrays arr1[0..n-1] and arr2[0..m-1]
def LCIS(arr1, n, arr2, m):
 
    # table[j] is going to store length of LCIS
    # ending with arr2[j]. We initialize it as 0,
    table = [0] * m
    for j in range(m):
        table[j] = 0
 
    # Traverse all elements of arr1[]
    for i in range(n):
     
        # Initialize current length of LCIS
        current = 0
 
        # For each element of arr1[],
        # traverse all elements of arr2[].
        for j in range(m):
             
            # If both the array have same elements.
            # Note that we don't break the loop here.
            if (arr1[i] == arr2[j]):
                if (current + 1 > table[j]):
                    table[j] = current + 1
 
            # Now seek for previous smaller common
            # element for current element of arr1
            if (arr1[i] > arr2[j]):
                if (table[j] > current):
                    current = table[j]
 
    # The maximum value in table[]
    # is out result
    result = 0
    for i in range(m):
        if (table[i] > result):
            result = table[i]
 
    return result
 
# Driver Code
if __name__ == "__main__":
     
    arr1 = [3, 4, 9, 1]
    arr2 = [5, 3, 8, 9, 10, 2, 1]
 
    n = len(arr1)
    m = len(arr2)
 
    print("Length of LCIS is",
           LCIS(arr1, n, arr2, m))
 
# This code is contributed by ita_c

C#

// A C# Program to find length of the Longest
// Common Increasing Subsequence (LCIS)
using System;
 
class GFG {
 
    // Returns the length and the LCIS of two
    // arrays arr1[0..n-1] and arr2[0..m-1]
    static int LCIS(int []arr1, int n,
                    int []arr2, int m)
    {
        // table[j] is going to store length of
        // LCIS ending with arr2[j]. We initialize
        // it as 0,
        int []table = new int[m];
        for (int j = 0; j < m; j++)
            table[j] = 0;
 
        // Traverse all elements of arr1[]
        for (int i = 0; i < n; i++)
        {
            // Initialize current length of LCIS
            int current = 0;
 
            // For each element of arr1[], trvarse
            // all elements of arr2[].
            for (int j = 0; j < m; j++)
            {
                // If both the array have same
                // elements. Note that we don't
                // break the loop here.
                if (arr1[i] == arr2[j])
                    if (current + 1 > table[j])
                        table[j] = current + 1;
 
                /* Now seek for previous smaller
                   common element for current
                   element of arr1 */
                if (arr1[i] > arr2[j])
                    if (table[j] > current)
                        current = table[j];
            }
        }
 
        // The maximum value in
        // table[] is out result
        int result = 0;
        for (int i = 0; i < m; i++)
            if (table[i] > result)
            result = table[i];
 
        return result;
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int []arr1 = {3, 4, 9, 1};
        int []arr2 = {5, 3, 8, 9, 10, 2, 1};
 
        int n = arr1.Length;
        int m = arr2.Length;
 
    Console.Write("Length of LCIS is " +
                   LCIS(arr1, n, arr2, m));
    }
}
 
// This code is contributed by nitin mittal.

PHP

 $table[$j])
                    $table[$j] = $current + 1;
 
            /* Now seek for previous smaller
            common element for current
            element of arr1 */
            if ($arr1[$i] > $arr2[$j])
                if ($table[$j] > $current)
                    $current = $table[$j];
        }
    }
 
    // The maximum value in
    // table[] is out result
    $result = 0;
    for ($i = 0; $i < $m; $i++)
        if ($table[$i] > $result)
        $result = $table[$i];
 
    return $result;
}
 
// Driver Code
$arr1 = array (3, 4, 9, 1);
$arr2 = array (5, 3, 8, 9, 10, 2, 1);
 
$n = sizeof($arr1);
$m = sizeof($arr2);
 
echo "Length of LCIS is ",
       LCIS($arr1, $n, $arr2, $m);
 
// This code is contributed by ajit
?>

Javascript

 
// Javascript Program to find length of the Longest
// Common Increasing Subsequence (LCIS)
 
    // Returns the length and the LCIS of two
    // arrays arr1[0..n-1] and arr2[0..m-1]
    function LCIS(arr1, n, arr2, m)
    {
        // table[j] is going to store length of
        // LCIS ending with arr2[j]. We initialize
        // it as 0,
        let table = [];
        for (let j = 0; j < m; j++)
            table[j] = 0;
   
        // Traverse all elements of arr1[]
        for (let i = 0; i < n; i++)
        {
            // Initialize current length of LCIS
            let current = 0;
   
            // For each element of arr1[], trvarse
            // all elements of arr2[].
            for (let j = 0; j < m; j++)
            {
                // If both the array have same
                // elements. Note that we don't
                // break the loop here.
                if (arr1[i] == arr2[j])
                    if (current + 1 > table[j])
                        table[j] = current + 1;
   
                /* Now seek for previous smaller
                common element for current
                element of arr1 */
                if (arr1[i] > arr2[j])
                    if (table[j] > current)
                        current = table[j];
            }
        }
   
        // The maximum value in table[] is out
        // result
        let result = 0;
        for (let i=0; i<m; i++)
            if (table[i] > result)
            result = table[i];
   
        return result;
    }
   
 
// Driver Code
 
        let arr1 = [3, 4, 9, 1];
        let arr2 = [5, 3, 8, 9, 10, 2, 1];
   
        let n = arr1.length;
        let m = arr2.length;
   
    document.write("Length of LCIS is " +
                       LCIS(arr1, n, arr2, m));
                       

输出 ：

Length of LCIS is 2