題目

https://leetcode.com/problems/maximum-length-of-repeated-subarray/

題解

Dynamic Programming [Accepted]

參考：官方題解

Intuition and Algorithm

Since a common subarray of A and B must start at some A[i] and B[j], let dp[i][j] be the longest common prefix of A[i:] and B[j:]. Whenever A[i] == B[j], we know dp[i][j] = dp[i+1][j+1] + 1. Also, the answer is max(dp[i][j]) over all i, j.

We can perform bottom-up dynamic programming to find the answer based on this recurrence. Our loop invariant is that the answer is already calculated correctly and stored in dp for any larger i, j.

class Solution {public int findLength(int[] nums1, int[] nums2) {int[][] same = new int[nums1.length][nums2.length];int max = 0;for (int i = 0; i < nums1.length; i++) {for (int j = 0; j < nums2.length; j++) {if (nums1[i] == nums2[j]) {same[i][j] = 1;if (i > 0 && j > 0) same[i][j] += same[i - 1][j - 1];max = Math.max(max, same[i][j]);}}}return max;} }

總結

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