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‎README.adoc

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//|{doc_base_url}/0717-1-bit-and-2-bit-characters.adoc[题解]
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//|Easy
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//|
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//
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//|{counter:codes}
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//|{leetcode_base_url}/maximum-length-of-repeated-subarray/[718. Maximum Length of Repeated Subarray^]
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//|{source_base_url}/_0718_MaximumLengthOfRepeatedSubarray.java[Java]
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//|{doc_base_url}/0718-maximum-length-of-repeated-subarray.adoc[题解]
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//|Medium
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/maximum-length-of-repeated-subarray/[718. Maximum Length of Repeated Subarray^]
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|{source_base_url}/_0718_MaximumLengthOfRepeatedSubarray.java[Java]
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|{doc_base_url}/0718-maximum-length-of-repeated-subarray.adoc[题解]
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|Medium
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|
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//|{counter:codes}
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//|{leetcode_base_url}/find-k-th-smallest-pair-distance/[719. Find K-th Smallest Pair Distance^]
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//|{source_base_url}/_0719_FindKThSmallestPairDistance.java[Java]

‎docs/0000-26-dp-5-longest-common-substring.adoc

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. xref:0300-longest-increasing-subsequence.adoc[300. Longest Increasing Subsequence]
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. xref:1143-longest-common-subsequence.adoc[1143. Longest Common Subsequence]
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. xref:1092-shortest-common-supersequence.adoc[1092. Shortest Common Supersequence]
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. xref:0718-maximum-length-of-repeated-subarray.adoc[718. 最长重复子数组]
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Longest Common Substring，最长相同子串
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‎docs/0718-maximum-length-of-repeated-subarray.adoc

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[#0718-maximum-length-of-repeated-subarray]
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= 718. Maximum Length of Repeated Subarray
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= 718. 最长重复子数组
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{leetcode}/problems/maximum-length-of-repeated-subarray/[LeetCode - Maximum Length of Repeated Subarray^]
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https://leetcode.cn/problems/maximum-length-of-repeated-subarray/[LeetCode - 718. 最长重复子数组 ^]
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Given two integer arrays `A` and `B`, return the maximum length of an subarray that appears in both arrays.
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给两个整数数组 `+nums1+` 和 `+nums2+` ，返回 _两个数组中 *公共的*
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、长度最长的子数组的长度 _。
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*Example 1:*
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*示例 1：*
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[subs="verbatim,quotes,macros"]
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----
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*Input:*
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A: [1,2,3,2,1]
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B: [3,2,1,4,7]
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*Output:* 3
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*Explanation:*
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The repeated subarray with maximum length is [3, 2, 1].
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----
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....
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输入：nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
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输出：3
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解释：长度最长的公共子数组是 [3,2,1] 。
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....
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*示例 2：*
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....
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输入：nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
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输出：5
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....
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*Note:*
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*提示：*
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* `+1 <= nums1.length, nums2.length <= 1000+`
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* `+0 <= nums1[i], nums2[i] <= 100+`
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. 1 <= len(A), len(B) <= 1000
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. 0 <= A[i], B[i] < 100
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== 思路分析
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动态规划：
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image::images/0718-01.png[{image_attr}]
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image::images/0718-02.png[{image_attr}]
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image::images/0718-03.png[{image_attr}]
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滑动窗口的解法非常妙！
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image:images/0718-01.gif[{image_attr}]
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[[src-0718]]
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[tabs]
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====
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一刷::
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+
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0718_MaximumLengthOfRepeatedSubarray.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0718_MaximumLengthOfRepeatedSubarray_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/maximum-length-of-repeated-subarray/solutions/310099/zui-chang-zhong-fu-zi-shu-zu-by-leetcode-solution/[718. 最长重复子数组 - 官方题解^]
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. https://leetcode.cn/problems/maximum-length-of-repeated-subarray/solutions/28583/wu-li-jie-fa-by-stg-2/[718. 最长重复子数组 - 滑动窗口解法^]
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. https://leetcode.cn/problems/maximum-length-of-repeated-subarray/solutions/310509/zhe-yao-jie-shi-ken-ding-jiu-dong-liao-by-hyj8/[718. 最长重复子数组 -「手画图解」动态规划 思路解析^]
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‎docs/images/0718-01.gif

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‎docs/images/0718-01.png

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‎docs/images/0718-02.png

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‎docs/images/0718-03.png

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‎docs/index.adoc

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// include::0717-1-bit-and-2-bit-characters.adoc[leveloffset=+1]
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// include::0718-maximum-length-of-repeated-subarray.adoc[leveloffset=+1]
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include::0718-maximum-length-of-repeated-subarray.adoc[leveloffset=+1]
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// include::0719-find-k-th-smallest-pair-distance.adoc[leveloffset=+1]
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‎logbook/202406.adoc

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|{doc_base_url}/0072-edit-distance.adoc[题解]
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|❌ 动态规划，看答案秒懂，自己想不出来！
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|{counter:codes}
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|{leetcode_base_url}/maximum-length-of-repeated-subarray/[718. Maximum Length of Repeated Subarray^]
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|{doc_base_url}/0718-maximum-length-of-repeated-subarray.adoc[题解]
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|❌ 动态规划。滑动窗口的解法也很妙！
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|===
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‎src/main/java/com/diguage/algo/leetcode/_0718_MaximumLengthOfRepeatedSubarray.java

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package com.diguage.algo.leetcode;
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public class _0718_MaximumLengthOfRepeatedSubarray {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2024-09-30 20:18:46
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*/
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public int findLength(int[] nums1, int[] nums2) {
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int n = nums1.length, m = nums2.length;
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int[][] dp = new int[n + 1][m + 1];
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int result = 0;
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for (int i = n - 1; i >= 0; i--) {
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for (int j = m - 1; j >= 0; j--) {
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dp[i][j] = nums1[i] == nums2[j] ? dp[i + 1][j + 1] + 1 : 0;
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result = Math.max(result, dp[i][j]);
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}
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}
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return result;
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}
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// end::answer[]
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}

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