给定两个数组arr1 []和arr2 [] ，任务是找到arr1 []的最长子数组，它是arr2 []的子序列。
例子：
Input: arr1[] = {4, 2, 3, 1, 5, 6}, arr2[] = {3, 1, 4, 6, 5, 2}
Output: 3
Explanation: The longest subarray of arr1[] which is a subsequence in arr2[] is {3, 1, 5}
Input: arr1[] = {3, 2, 4, 7, 1, 5, 6, 8, 10, 9}, arr2[] = {9, 2, 4, 3, 1, 5, 6, 8, 10, 7}
Output: 5
Explanation: The longest subarray in arr1[] which is a subsequence in arr2[] is {1, 5, 6, 8, 10}.
方法：想法是使用动态编程来解决此问题。请按照以下步骤解决问题：
- 初始化DP [] []表，其中DP [i] [j]存储直到arr1 []中第i个索引的最长子数组的长度，它是arr2 []中第j个索引的子序列。
- 现在，遍历两个数组并执行以下操作：
- 情况1：如果arr1 [i]和arr2 [j]相等，则将1加到DP [i-1] [j-1]，因为arr1 [i]和arr2 [j]有助于最长子数组的所需长度。
- 情况2：如果arr1 [i]和arr2 [j]不相等，则设置DP [i] [j] = DP [i – 1] [j] 。
- 最后，将DP [] []表中存在的最大值打印为所需答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the length of the
// longest subarray in arr1[] which
// is a subsequence in arr2[]
int LongSubarrSeq(int arr1[], int arr2[], int M, int N)
{
// Length of the array arr1[]
// Length of the required
// longest subarray
int maxL = 0;
// Initialize DP[]array
int DP[M + 1][N + 1];
// Traverse array arr1[]
for (int i = 1; i <= M; i++)
{
// Traverse array arr2[]
for (int j = 1; j <= N; j++)
{
if (arr1[i - 1] == arr2[j - 1])
{
// arr1[i - 1] contributes to
// the length of the subarray
DP[i][j] = 1 + DP[i - 1][j - 1];
}
// Otherwise
else
{
DP[i][j] = DP[i][j - 1];
}
}
}
// Find the maximum value
// present in DP[][]
for (int i = 1; i <= M; i++)
{
for (int j = 1; j <= N; j++)
{
maxL = max(maxL, DP[i][j]);
}
}
// Return the result
return maxL;
}
// Driver Code
int main()
{
int arr1[] = { 4, 2, 3, 1, 5, 6 };
int M = sizeof(arr1) / sizeof(arr1[0]);
int arr2[] = { 3, 1, 4, 6, 5, 2 };
int N = sizeof(arr2) / sizeof(arr2[0]);
// Function call to find the length
// of the longest required subarray
cout << LongSubarrSeq(arr1, arr2, M, N) < Java
// Java program
// for the above approach
import java.io.*;
class GFG {
// Function to find the length of the
// longest subarray in arr1[] which
// is a subsequence in arr2[]
private static int LongSubarrSeq(
int[] arr1, int[] arr2)
{
// Length of the array arr1[]
int M = arr1.length;
// Length of the array arr2[]
int N = arr2.length;
// Length of the required
// longest subarray
int maxL = 0;
// Initialize DP[]array
int[][] DP = new int[M + 1][N + 1];
// Traverse array arr1[]
for (int i = 1; i <= M; i++) {
// Traverse array arr2[]
for (int j = 1; j <= N; j++) {
if (arr1[i - 1] == arr2[j - 1]) {
// arr1[i - 1] contributes to
// the length of the subarray
DP[i][j] = 1 + DP[i - 1][j - 1];
}
// Otherwise
else {
DP[i][j] = DP[i][j - 1];
}
}
}
// Find the maximum value
// present in DP[][]
for (int i = 1; i <= M; i++) {
for (int j = 1; j <= N; j++) {
maxL = Math.max(maxL, DP[i][j]);
}
}
// Return the result
return maxL;
}
// Driver Code
public static void main(String[] args)
{
int[] arr1 = { 4, 2, 3, 1, 5, 6 };
int[] arr2 = { 3, 1, 4, 6, 5, 2 };
// Function call to find the length
// of the longest required subarray
System.out.println(LongSubarrSeq(arr1, arr2));
}
}Python3
# Python program
# for the above approach
# Function to find the length of the
# longest subarray in arr1 which
# is a subsequence in arr2
def LongSubarrSeq(arr1, arr2):
# Length of the array arr1
M = len(arr1);
# Length of the array arr2
N = len(arr2);
# Length of the required
# longest subarray
maxL = 0;
# Initialize DParray
DP = [[0 for i in range(N + 1)] for j in range(M + 1)];
# Traverse array arr1
for i in range(1, M + 1):
# Traverse array arr2
for j in range(1, N + 1):
if (arr1[i - 1] == arr2[j - 1]):
# arr1[i - 1] contributes to
# the length of the subarray
DP[i][j] = 1 + DP[i - 1][j - 1];
# Otherwise
else:
DP[i][j] = DP[i][j - 1];
# Find the maximum value
# present in DP
for i in range(M + 1):
# Traverse array arr2
for j in range(1, N + 1):
maxL = max(maxL, DP[i][j]);
# Return the result
return maxL;
# Driver Code
if __name__ == '__main__':
arr1 = [4, 2, 3, 1, 5, 6];
arr2 = [3, 1, 4, 6, 5, 2];
# Function call to find the length
# of the longest required subarray
print(LongSubarrSeq(arr1, arr2));
# This code contributed by shikhasingrajputC#
// C# program for the above approach
using System;
class GFG{
// Function to find the length of the
// longest subarray in arr1[] which
// is a subsequence in arr2[]
private static int LongSubarrSeq(int[] arr1,
int[] arr2)
{
// Length of the array arr1[]
int M = arr1.Length;
// Length of the array arr2[]
int N = arr2.Length;
// Length of the required
// longest subarray
int maxL = 0;
// Initialize DP[]array
int[,] DP = new int[M + 1, N + 1];
// Traverse array arr1[]
for(int i = 1; i <= M; i++)
{
// Traverse array arr2[]
for(int j = 1; j <= N; j++)
{
if (arr1[i - 1] == arr2[j - 1])
{
// arr1[i - 1] contributes to
// the length of the subarray
DP[i, j] = 1 + DP[i - 1, j - 1];
}
// Otherwise
else
{
DP[i, j] = DP[i, j - 1];
}
}
}
// Find the maximum value
// present in DP[][]
for(int i = 1; i <= M; i++)
{
for(int j = 1; j <= N; j++)
{
maxL = Math.Max(maxL, DP[i, j]);
}
}
// Return the result
return maxL;
}
// Driver Code
static public void Main()
{
int[] arr1 = { 4, 2, 3, 1, 5, 6 };
int[] arr2 = { 3, 1, 4, 6, 5, 2 };
// Function call to find the length
// of the longest required subarray
Console.WriteLine(LongSubarrSeq(arr1, arr2));
}
}
// This code is contributed by susmitakundugoaldanga输出：
3时间复杂度： O(M * N)
辅助空间： O(M * N)