1. subject

Give two arrays of integers nums1 and nums2 , Returns the common of two arrays 、 The length of the longest subarray .

Example 1：
Input ：nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
Output ：3
explain ： The longest common subarray is [3,2,1] .

Example 2：
Input ：nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
Output ：5

Tips ：
1 <= nums1.length, nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 100

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/maximum-length-of-repeated-subarray

2. Ideas

（1） The law of violent exhaustion
This method is easy to think of , namely Exhaustive array nums1 Starting position in i And an array nums2 Starting position in j, And then calculate nums1[i…length1 - 1] and nums2[j…length2 - 1] The longest public prefix of k, Use variables in the calculation process res Record k The maximum of , Return after exhausting res that will do . The time complexity of this method is O(n³).

（2） Dynamic programming
Train of thought reference Official solution to this problem .

3. Code implementation （Java）

// Ideas 1———— The law of violent exhaustion 
class Solution {
    
    public int findLength(int[] nums1, int[] nums2) {
    
        int res = 0;
        int length1 = nums1.length;
        int length2 = nums2.length;
        for (int i = 0; i < length1; i++) {
    
            for (int j = 0; j < length2; j++) {
    
                int k = 0;
                while (i + k < length1 && j + k <length2 && nums1[i + k] == nums2[j + k]) {
    
                    k++;
                }
                res = Math.max(res, k);
            }
        }
        return res;
    }
}

// Ideas 2———— Dynamic programming 
class Solution {
    
    public int findLength(int[] nums1, int[] nums2) {
    
        int res = 0;
        int length1 = nums1.length;
        int length2 = nums2.length;
        //dp[i][j]  Express  nums1[i...length1 - 1]  and  nums2[j...length2 - 1]  The longest public prefix of 
        int[][] dp = new int[length1 + 1][length2 + 1];
        for (int i = length1 - 1; i >= 0; i--) {
    
            for (int j = length2 - 1; j >=0; j--) {
    
                dp[i][j] = (nums1[i] == nums2[j]) ? dp[i + 1][j + 1] + 1 : 0;
                res = Math.max(dp[i][j], res);
            }
        }
        return res;
    }
}