diff --git a/Algorithms/0718. Maximum Length of Repeated Subarray/718. Maximum Length of Repeated Subarray.go b/Algorithms/0718. Maximum Length of Repeated Subarray/718. Maximum Length of Repeated Subarray.go
new file mode 100644
index 00000000..3eec755f
--- /dev/null
+++ b/Algorithms/0718. Maximum Length of Repeated Subarray/718. Maximum Length of Repeated Subarray.go	
@@ -0,0 +1,79 @@
+package leetcode
+
+const primeRK = 16777619
+
+// 解法一 二分搜索 + Rabin-Karp
+func findLength(A []int, B []int) int {
+	low, high := 0, min(len(A), len(B))
+	for low < high {
+		mid := (low + high + 1) >> 1
+		if hasRepeated(A, B, mid) {
+			low = mid
+		} else {
+			high = mid - 1
+		}
+	}
+	return low
+}
+
+func hashSlice(arr []int, length int) []int {
+	// hash 数组里面记录 arr 比 length 长出去部分的 hash 值
+	hash, pl, h := make([]int, len(arr)-length+1), 1, 0
+	for i := 0; i < length-1; i++ {
+		pl *= primeRK
+	}
+	for i, v := range arr {
+		h = h*primeRK + v
+		if i >= length-1 {
+			hash[i-length+1] = h
+			h -= pl * arr[i-length+1]
+		}
+	}
+	return hash
+}
+
+func hasSamePrefix(A, B []int, length int) bool {
+	for i := 0; i < length; i++ {
+		if A[i] != B[i] {
+			return false
+		}
+	}
+	return true
+}
+
+func hasRepeated(A, B []int, length int) bool {
+	hs := hashSlice(A, length)
+	hashToOffset := make(map[int][]int, len(hs))
+	for i, h := range hs {
+		hashToOffset[h] = append(hashToOffset[h], i)
+	}
+	for i, h := range hashSlice(B, length) {
+		if offsets, ok := hashToOffset[h]; ok {
+			for _, offset := range offsets {
+				if hasSamePrefix(A[offset:], B[i:], length) {
+					return true
+				}
+			}
+		}
+	}
+	return false
+}
+
+// 解法二 DP 动态规划
+func findLength1(A []int, B []int) int {
+	res, dp := 0, make([][]int, len(A)+1)
+	for i := range dp {
+		dp[i] = make([]int, len(B)+1)
+	}
+	for i := len(A) - 1; i >= 0; i-- {
+		for j := len(B) - 1; j >= 0; j-- {
+			if A[i] == B[j] {
+				dp[i][j] = dp[i+1][j+1] + 1
+				if dp[i][j] > res {
+					res = dp[i][j]
+				}
+			}
+		}
+	}
+	return res
+}
diff --git a/Algorithms/0718. Maximum Length of Repeated Subarray/718. Maximum Length of Repeated Subarray_test.go b/Algorithms/0718. Maximum Length of Repeated Subarray/718. Maximum Length of Repeated Subarray_test.go
new file mode 100644
index 00000000..60bdddc2
--- /dev/null
+++ b/Algorithms/0718. Maximum Length of Repeated Subarray/718. Maximum Length of Repeated Subarray_test.go	
@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question718 struct {
+	para718
+	ans718
+}
+
+// para 是参数
+// one 代表第一个参数
+type para718 struct {
+	A []int
+	B []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans718 struct {
+	one int
+}
+
+func Test_Problem718(t *testing.T) {
+
+	qs := []question718{
+
+		question718{
+			para718{[]int{0, 0, 0, 0, 0}, []int{0, 0, 0, 0, 0}},
+			ans718{5},
+		},
+
+		question718{
+			para718{[]int{1, 2, 3, 2, 1}, []int{3, 2, 1, 4, 7}},
+			ans718{3},
+		},
+
+		question718{
+			para718{[]int{0, 0, 0, 0, 1}, []int{1, 0, 0, 0, 0}},
+			ans718{4},
+		},
+
+		question718{
+			para718{[]int{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, []int{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0}},
+			ans718{59},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 718------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans718, q.para718
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findLength(p.A, p.B))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0718. Maximum Length of Repeated Subarray/README.md b/Algorithms/0718. Maximum Length of Repeated Subarray/README.md
new file mode 100755
index 00000000..71b2388e
--- /dev/null
+++ b/Algorithms/0718. Maximum Length of Repeated Subarray/README.md	
@@ -0,0 +1,37 @@
+# [718. Maximum Length of Repeated Subarray](https://leetcode.com/problems/maximum-length-of-repeated-subarray/)
+
+
+## 题目:
+
+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. 1 <= len(A), len(B) <= 1000
+2. 0 <= A[i], B[i] < 100
+
+
+## 题目大意
+
+给两个整数数组 A 和 B ，返回两个数组中公共的、长度最长的子数组的长度。
+
+
+
+## 解题思路
+
+- 给出两个数组，求这两个数组中最长相同子串的长度。
+- 这一题最容易想到的是 DP 动态规划的解法。`dp[i][j]` 代表在 A 数组中以 `i` 下标开始的子串与 B 数组中以 `j` 下标开始的子串最长相同子串的长度，状态转移方程为 `dp[i][j] = dp[i+1][j+1] + 1` (当 `A[i] == B[j]`)。这种解法的时间复杂度是 O(n^2)，空间复杂度 O(n^2)。
+- 这一题最佳解法是二分搜索 + `Rabin-Karp`。比较相同子串耗时的地方在于，需要一层循环，遍历子串所有字符。但是如果比较两个数字就很快，`O(1)` 的时间复杂度。所以有人就想到了，能不能把字符串也映射成数字呢？这样比较起来就非常快。这个算法就是 `Rabin-Karp` 算法。字符串映射成一个数字不能随意映射，还要求能根据字符串前缀动态增加，比较下一个字符串的时候，可以利用已比较过的前缀，加速之后的字符串比较。在 Rabin-Karp 算法中有一个“码点”的概念。类似于10进制中的进制。具体的算法讲解可以见这篇：
+
+    [基础知识 - Rabin-Karp 算法](https://www.cnblogs.com/golove/p/3234673.html)
+
+    “码点”一般取值为一个素数。在 go 的 `strings` 包里面取值是 16777619。所以这一题也可以直接取这个值。由于这一次要求我们找最长长度，所以把最长长度作为二分搜索的目标。先将数组 A 和数组 B 中的数字都按照二分出来的长度，进行 `Rabin-Karp` hash。对 A 中的 hash 与下标做映射关系，存到 map 中，方便后面快速查找。然后遍历 B 中的 hash，当 hash 一致的时候，再匹配下标。如果下标存在，且拥有相同的前缀，那么就算找到了相同的子串了。最后就是不断的二分，找到最长的结果即可。这个解法的时间复杂度 O(n * log n)，空间复杂度 O(n)。
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