算法题目--Maximum Length of Repeated Subarray

题目来源

https://leetcode.com/problems/maximum-length-of-
4000
repeated-subarray/description/

题目描述

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

心路历程

最长回文子串，最长回文子序列，最长公共子串，最长公共子序列都是学习动态规划老生常谈的问题了。本题类似于最长公共子串，可以通过画二维矩阵的方法，找到最长的对角线来解决。

解决方案

#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
class Solution {
public:
int findLength(vector<int>& A, vector<int>& B) {
int lenA = A.size();
int lenB = B.size();
int max = 0;
int * temp = new int[lenA];
memset(temp, 0, sizeof(int)*lenA);//最开始写成sizeof(temp)报错
for (int i = 0; i < lenB; i++) {
for (int j = lenA - 1; j >= 0; j--) {
if (B[i] == A[j]) {
if (j == 0) {
temp[0] = 1;
if (max < temp[0]) max = temp[0];
}
else {
temp[j] = temp[j - 1] + 1;
if (max < temp[j]) max = temp[j];
}
}
else temp[j] = 0;
}
}
delete temp;
return max;
}
};
int main() {
vector<int> A = { 1,2,3,2,1 };
vector<int> B = { 3,2,1,4,7 };
Solution solution;
cout << solution.findLength(A, B)<<endl;
system("pause");
}