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) { int n1 = nums1.length, n2 = nums2.length; int[][] dp = new int[n1 + 1][n2 + 1]; int res = 0; for(int i = 0; i <= n1; i++) { for(int j = 0; j <= n2; j++) { if(i == 0 || j == 0) { dp[i][j] = 0; continue; } if(nums1[i - 1] == nums2[j - 1]) { dp[i][j] = 1 + dp[i - 1][j - 1]; res = Math.max(dp[i][j], res); } } } return res; } }
典中典dp用处之一---匹配