LeetCode Maximum Length of Repeated Subarray

This article introduces the Maximum Length of Repeated Subarray of LeetCode. I hope you will learn patiently.

Given two integer arrays A and B, return the length of the common and longest subarray in the two arrays.

Example 1:

输入:A: [1,2,3,2,1]
B: [3,2,1,4,7]输出: 3解释: 长度最长的公共子数组是 [3, 2, 1]。

Instructions:

1. 1

2. ##0

Solution, this is a classic dynamic programming algorithm, as follows:

public class MaxLengthRepeatedSubarray {
    //动态规划算法
    public static int findLength(int[] A, int[] B) {
        int aSize = A.length;
        int bSize = B.length;
        int[][] dp = new int[aSize + 1][bSize + 1];
        int result = 0;
        for (int i = 1; i < dp.length; i++) {
            for (int j = 1; j < dp[i].length; j++) {
                dp[i][j] = A[i - 1] == B[j - 1] ? dp[i - 1][j - 1] + 1 : 0;
                result = Math.max(result, dp[i][j]);
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] a = new int[]{1, 2, 3, 2, 1};
        int[] b = new int[]{3, 2, 1, 4, 7};
        System.out.println(findLength(a, b));
    }
}

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