maximum-length-of-repeated-subarray

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Algorithms

Given two integer arrays <code>nums1</code> and <code>nums2</code>, return the maximum length of a subarray that appears in both arrays.

&nbsp;
Example 1:

<pre>
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].
</pre>

Example 2:

<pre>
Input: nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
Output: 5
Explanation: The repeated subarray with maximum length is [0,0,0,0,0].
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 100</code></li>
</ul>

findLength

Maximum Length of Repeated Subarray

Minimum Size Subarray Sum

minimum-size-subarray-sum

长度最小的子数组

Longest Common Subpath

longest-common-subpath

最长公共子路径

Find the Maximum Length of a Good Subsequence II

find-the-maximum-length-of-a-good-subsequence-ii

求出最长好子序列 II

Find the Maximum Length of a Good Subsequence I

find-the-maximum-length-of-a-good-subsequence-i

求出最长好子序列 I

Array

Binary Search

Dynamic Programming

Sliding Window

Hash Function

Rolling Hash

给两个整数数组&nbsp;<code>nums1</code>&nbsp;和&nbsp;<code>nums2</code>&nbsp;，返回 两个数组中 公共的 、长度最长的子数组的长度&nbsp;。

&nbsp;

示例 1：

<pre>
输入：nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
输出：3
解释：长度最长的公共子数组是 [3,2,1] 。
</pre>

示例 2：

<pre>
输入：nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
输出：5
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 100</code></li>
</ul>