给定分别为N和M个整数的两个数组A []和B [] ，任务是找到两个给定数组之间相等的子数组的最大长度或最长的公共子数组。
例子：
Input: A[] = {1, 2, 8, 2, 1}, B[] = {8, 2, 1, 4, 7}
Output: 3
Explanation:
The subarray that is common to both arrays are {8, 2, 1} and the length of the subarray is 3.
Input: A[] = {1, 2, 3, 2, 1}, B[] = {8, 7, 6, 4, 7}
Output: 0
Explanation:
There is no such subarrays which are equal in the array A[] and B[].
天真的方法：想法是生成两个给定数组A []和B []的所有子数组，并找到最长的匹配子数组。就时间复杂度而言，此解决方案是指数级的。
时间复杂度： O(2 N + M )，其中N是数组A []的长度，M是数组B []的长度。
高效方法：
有效的方法是使用动态编程(DP)。此问题是最长公共子序列(LCS)的变体。
令输入序列分别为长度为m和n的A [0..n-1]和B [0..m-1] 。以下是equal子数组的递归实现：
- 由于A []和B []的公共子数组必须从某个索引i和j开始，使得A [i]等于B [j] 。令dp [i] [j]为A [i …]和B [j …]的最长公共子数组。
- 因此，对于任何索引i和j，如果A [i]等于B [j]，则dp [i] [j] = dp [i + 1] [j + 1] +1 。
- 数组dp [] []中所有元素的最大值将给出相等子数组的最大长度。
例如：
如果给定数组A [] = {1,2,8,2,1}和B [] = {8，2，1，4，7}。如果字符在数组A []和B []的索引i和j处分别匹配，则dp [i] [j]将更新为1 + dp [i + 1] [j + 1] 。
以下是给定数组A []和B []的更新后的dp [] []表。
下面是上述方法的实现：
C++
// C++ program to DP approach
// to above solution
#include
using namespace std;
// Function to find the maximum
// length of equal subarray
int FindMaxLength(int A[], int B[], int n, int m)
{
// Auxillary dp[][] array
int dp[n + 1][m + 1];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
dp[i][j] = 0;
// Updating the dp[][] table
// in Bottom Up approach
for (int i = n - 1; i >= 0; i--)
{
for (int j = m - 1; j >= 0; j--)
{
// If A[i] is equal to B[i]
// then dp[j][i] = dp[j + 1][i + 1] + 1
if (A[i] == B[j])
dp[j][i] = dp[j + 1][i + 1] + 1;
}
}
int maxm = 0;
// Find maximum of all the values
// in dp[][] array to get the
// maximum length
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Update the length
maxm = max(maxm, dp[i][j]);
}
}
// Return the maximum length
return maxm;
}
// Driver Code
int main()
{
int A[] = { 1, 2, 8, 2, 1 };
int B[] = { 8, 2, 1, 4, 7 };
int n = sizeof(A) / sizeof(A[0]);
int m = sizeof(B) / sizeof(B[0]);
// Function call to find
// maximum length of subarray
cout << (FindMaxLength(A, B, n, m));
}
// This code is contributed by chitranayal Java
// Java program to DP approach
// to above solution
class GFG
{
// Function to find the maximum
// length of equal subarray
static int FindMaxLength(int A[], int B[], int n, int m)
{
// Auxillary dp[][] array
int[][] dp = new int[n + 1][m + 1];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
dp[i][j] = 0;
// Updating the dp[][] table
// in Bottom Up approach
for (int i = n - 1; i >= 0; i--)
{
for (int j = m - 1; j >= 0; j--)
{
// If A[i] is equal to B[i]
// then dp[j][i] = dp[j + 1][i + 1] + 1
if (A[i] == B[j])
dp[j][i] = dp[j + 1][i + 1] + 1;
}
}
int maxm = 0;
// Find maximum of all the values
// in dp[][] array to get the
// maximum length
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Update the length
maxm = Math.max(maxm, dp[i][j]);
}
}
// Return the maximum length
return maxm;
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 1, 2, 8, 2, 1 };
int B[] = { 8, 2, 1, 4, 7 };
int n = A.length;
int m = B.length;
// Function call to find
// maximum length of subarray
System.out.print(FindMaxLength(A, B, n, m));
}
}
// This code is contributed by PrinciRaj1992Python
# Python program to DP approach
# to above solution
# Function to find the maximum
# length of equal subarray
def FindMaxLength(A, B):
n = len(A)
m = len(B)
# Auxillary dp[][] array
dp = [[0 for i in range(n + 1)] for i in range(m + 1)]
# Updating the dp[][] table
# in Bottom Up approach
for i in range(n - 1, -1, -1):
for j in range(m - 1, -1, -1):
# If A[i] is equal to B[i]
# then dp[j][i] = dp[j + 1][i + 1] + 1
if A[i] == B[j]:
dp[j][i] = dp[j + 1][i + 1] + 1
maxm = 0
# Find maximum of all the values
# in dp[][] array to get the
# maximum length
for i in dp:
for j in i:
# Update the length
maxm = max(maxm, j)
# Return the maximum length
return maxm
# Driver Code
if __name__ == '__main__':
A = [1, 2, 8, 2, 1]
B = [8, 2, 1, 4, 7]
# Function call to find
# maximum length of subarray
print(FindMaxLength(A, B))C#
// C# program to DP approach
// to above solution
using System;
class GFG
{
// Function to find the maximum
// length of equal subarray
static int FindMaxLength(int[] A, int[] B, int n, int m)
{
// Auxillary [,]dp array
int[, ] dp = new int[n + 1, m + 1];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
dp[i, j] = 0;
// Updating the [,]dp table
// in Bottom Up approach
for (int i = n - 1; i >= 0; i--)
{
for (int j = m - 1; j >= 0; j--)
{
// If A[i] is equal to B[i]
// then dp[j, i] = dp[j + 1, i + 1] + 1
if (A[i] == B[j])
dp[j, i] = dp[j + 1, i + 1] + 1;
}
}
int maxm = 0;
// Find maximum of all the values
// in [,]dp array to get the
// maximum length
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Update the length
maxm = Math.Max(maxm, dp[i, j]);
}
}
// Return the maximum length
return maxm;
}
// Driver Code
public static void Main(String[] args)
{
int[] A = { 1, 2, 8, 2, 1 };
int[] B = { 8, 2, 1, 4, 7 };
int n = A.Length;
int m = B.Length;
// Function call to find
// maximum length of subarray
Console.Write(FindMaxLength(A, B, n, m));
}
}
// This code is contributed by PrinciRaj1992输出
3时间复杂度： O(N * M)，其中N是数组A []的长度，M是数组B []的长度。