From 219d1d9982bd57dd1249abc1c08f049215104ca1 Mon Sep 17 00:00:00 2001
From: hk27xing <244798299@qq.com>
Date: Sat, 10 Jul 2021 21:09:42 +0800
Subject: [PATCH] =?UTF-8?q?=E6=94=B9=E8=BF=9B718.=E6=9C=80=E9=95=BF?=
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---
 problems/0718.最长重复子数组.md | 25 +++++++++++++++++++++++--
 1 file changed, 23 insertions(+), 2 deletions(-)

diff --git a/problems/0718.最长重复子数组.md b/problems/0718.最长重复子数组.md
index bef616d3..9ee94a3d 100644
--- a/problems/0718.最长重复子数组.md
+++ b/problems/0718.最长重复子数组.md
@@ -20,7 +20,7 @@ B: [3,2,1,4,7]
 输出：3
 解释：
 长度最长的公共子数组是 [3, 2, 1] 。
- 
+
 提示：
 
 * 1 <= len(A), len(B) <= 1000
@@ -155,6 +155,7 @@ public:
 
 Java：
 ```java
+// 版本一
 class Solution {
     public int findLength(int[] nums1, int[] nums2) {
         int result = 0;
@@ -164,7 +165,7 @@ class Solution {
             for (int j = 1; j < nums2.length + 1; j++) {
                 if (nums1[i - 1] == nums2[j - 1]) {
                     dp[i][j] = dp[i - 1][j - 1] + 1;
-                    max = Math.max(max, dp[i][j]);
+                    result = Math.max(result, dp[i][j]);
                 }
             }
         }
@@ -172,6 +173,26 @@ class Solution {
         return result;
     }
 }
+
+// 版本二: 滚动数组
+class Solution {
+    public int findLength(int[] nums1, int[] nums2) {
+        int[] dp = new int[nums2.length + 1];
+        int result = 0;
+
+        for (int i = 1; i <= nums1.length; i++) {
+            for (int j = nums2.length; j > 0; j--) {
+                if (nums1[i - 1] == nums2[j - 1]) {
+                    dp[j] = dp[j - 1] + 1;
+                } else {
+                    dp[j] = 0;
+                }
+                result = Math.max(result, dp[j]);
+            }
+        }
+        return result;
+    }
+}
 ```
 
 Python：