Maximum Length of Repeated Subarray(算法分析week12)

Maximum Length of Repeated Subarray

题目来源：https://leetcode.com/problemset/algorithms/

-题目描述-
-解题思路-
-代码实现-

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

解题思路

题目大意是找到A、B两个数组中相同的子数组的最大长度。
定义二维数组length[A.size()+1][B.size()+1]，length[i][j]表示A与B中相同子数组(A中此子数组最后一位下标为i-1,B中此子数组最后一位下标为j-1)的长度。
length[i][j]=length[i-1][j-1]+1 , if A[i-1]==B[j-1];
两个数组中相同的子数组的最大长度就是length[][]中的最大值。

代码实现

@requires_authorizationclass Solution {public:    int findLength(vector<int>& A, vector<int>& B) {    int max = 0;    int **length = new int*[A.size() + 1];    for (int i = 0; i <= A.size(); i++) {        length[i] = new int[B.size() + 1];    }    for (int i = 0; i <= A.size(); i++) {        for (int j = 0; j <= B.size(); j++) {            length[i][j] = 0;        }    }    for (int i = 1; i <= A.size(); i++) {        for (int j = 1; j <= B.size(); j++) {            if (A[i - 1] == B[j - 1]) {                length[i][j] = 1 + length[i - 1][j - 1];            }            if (length[i][j] > max) max = length[i][j];        }    }    return max;}};