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
就是类似找最长公共子序列的题目,利用动态规划算法。找出对大的那个值,复杂度为o(n^2)。递推式:
if (A[i-1] == B[j-1])
match[i][j] = match[i - 1][j - 1] + 1;
class Solution {
public:
int findLength(vector<int>& A, vector<int>& B) {
int match[1001][1001] = {0};
int max = 0;
for (int i = 0; i <= A.size(); i++) {
for (int j = 0; j <= B.size(); j++) {
if (i == 0 || j == 0) {
match[i][j] = 0;
}
else {
if (A[i-1] == B[j-1]) {
match[i][j] = match[i - 1][j - 1] + 1;
if (match[i][j] > max) {
max = match[i][j];
}
}
}
}
}
return max;
}
};