From d72bd072aff9489e3afb08b46ad607cab3b36126 Mon Sep 17 00:00:00 2001
From: yuhuacao <caoyhseu@126.com>
Date: Thu, 6 Dec 2018 21:36:42 +0800
Subject: [PATCH] update

---
 .../ Maximum Length of Repeated Subarray.py   | 36 +++++++++++++++++++
 yuhuacao/readme.md                            |  1 +
 2 files changed, 37 insertions(+)
 create mode 100644 yuhuacao/ Maximum Length of Repeated Subarray.py

diff --git a/yuhuacao/ Maximum Length of Repeated Subarray.py b/yuhuacao/ Maximum Length of Repeated Subarray.py
new file mode 100644
index 0000000..0a71252
--- /dev/null
+++ b/yuhuacao/ Maximum Length of Repeated Subarray.py	
@@ -0,0 +1,36 @@
+"""
+Given two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.
+
+Example 1:
+
+Input:
+A: [1,2,3,2,1]
+B: [3,2,1,4,7]
+Output: 3
+
+Explanation: 
+The repeated subarray with maximum length is [3, 2, 1].
+
+Note:
+    1 <= len(A), len(B) <= 1000
+    0 <= A[i], B[i] < 100
+"""
+"""
+解体思路：采用动态规划的方法，维护矩阵DP，DP[i][j]代表以A[i-1]与B[j-1]结尾的公共字串的长度，公共字串必须以A[i-1]，B[j-1]结束，即当A[i-1] == B[j-1]时，DP[i][j] = DP[i-1][j-1] + 1; 当A[i-1] != B[j-1]时，以A[i-1]和B[j-1]结尾的公共字串长度为0,DP[i][j] = 0。输出最大的公共字串的长度即为最长重复字串。 
+"""
+
+class Solution(object): 
+	def findLength(self, A, B): 
+		"""
+    	:type A: List[int]
+        :type B: List[int]
+        :rtype: int
+        """ 
+		n1,n2 = len(A),len(B) 
+		dp = [[0 for _ in range(n2+1)] for _ in range(n1+1)] 
+		for i in range(1,n1+1): 
+			for j in range(1,n2+1): 
+				if A[i-1] == B[j-1]: 
+					dp[i][j] = dp[i-1][j-1] + 1 
+		return max(max(row) for row in dp)
+
diff --git a/yuhuacao/readme.md b/yuhuacao/readme.md
index 766f250..882fb90 100644
--- a/yuhuacao/readme.md
+++ b/yuhuacao/readme.md
@@ -1,3 +1,4 @@
 # yuhuacao贡献内容
 
 1. Merge Sorted Array
+2. Maximum Length of Repeated Subarray