718. Maximum Length of Repeated Subarray
Given two integer arrays nums1 and nums2, return the maximum length of a subarray that appears in both arrays.
Example 1:
Input: nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
Output: 3
Explanation: The repeated subarray with maximum length is [3,2,1].
Example 2:
Input: nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
Output: 5
Constraints:
- 1 <= nums1.length, nums2.length <= 1000
- 0 <= nums1[i], nums2[i] <= 100
class Solution {
public int findLength(int[] nums1, int[] nums2) {
if(nums1 == null||nums2 == null) return 0;
int m = nums1.length;
int n = nums2.length;
int max = 0;
//dp[i][j] is the length of longest common subarray ending with nums[i] and nums[j]
int[][] dp = new int[m + 1][n + 1];
for(int i = 0;i <= m;i++){
for(int j = 0;j <= n;j++){
if(i == 0 || j == 0){
dp[i][j] = 0;
}else{
if(nums1[i - 1] == nums2[j - 1]){
dp[i][j] = 1 + dp[i - 1][j - 1];
max = Math.max(max,dp[i][j]);
}
}
}
}
return max;
}
}