Given two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.

Example 1:
Input:
A: [1,2,3,2,1]
B: [3,2,1,4,7]
Output: 3
Explanation:
The repeated subarray with maximum length is [3, 2, 1].
Note:
1 <= len(A), len(B) <= 1000
0 <= A[i], B[i] < 100

本題就是尋找最長的公共子串

程式碼如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
 

#include <numeric>
#include <cmath>
#include <regex>

using namespace std;



class Solution 
{
public:
    int findLength(vector<int>& a, vector<int>& b) 
    {
        int m = a.size(), n = b.size();
        vector<vector<int>> dp(m+1,vector<int
 
>(n+1,0));

        int res = 0;
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (a[i - 1] == b[j - 1])
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                else
                    dp[i][j] = 0;
                res = max(res, dp[i][j]);
            }
        }
        return res;
    }
};