[LeetCode]Maximum Length of Repeated Subarray

Maximum Length of Repeated Subarray:

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

这是一道典型的动态规划题目。

寻找子序列的规律如下：



具体思想可见动态规划解最长公共子序列问题。

而这是一道找公共子串的题目，和上面不同的是，当xi != yi时，c[i,j]=0。

class Solution {
public:
int findLength(vector<int>& A, vector<int>& B) {
int temp[1001][1001],max = 0;
for(int i = 0; i < A.size(); i++){
for(int j = 0; j < B.size(); j++){
if(A[i] == B[j]){
temp[i+1][j+1] = temp[i][j] + 1;
}else{
temp[i+1][j+1] = 0;
}
max = max>temp[i+1][j+1]?max:temp[i+1][j+1];
}
}
return max;
}
};