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feat: add solutions to lc problem: No.0718
No.0718.Maximum Length of Repeated Subarray
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‎solution/0700-0799/0718.Maximum Length of Repeated Subarray/README.md

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Original file line numberDiff line numberDiff line change
@@ -38,22 +38,99 @@
3838

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def findLength(self, nums1: List[int], nums2: List[int]) -> int:
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m, n = len(nums1), len(nums2)
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dp = [[0] * (n + 1) for _ in range(m + 1)]
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ans = 0
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for i in range(1, m + 1):
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for j in range(1, n + 1):
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if nums1[i - 1] == nums2[j - 1]:
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dp[i][j] = 1 + dp[i - 1][j - 1]
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ans = max(ans, dp[i][j])
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int findLength(int[] nums1, int[] nums2) {
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int m = nums1.length;
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int n = nums2.length;
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int[][] dp = new int[m + 1][n + 1];
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int ans = 0;
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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if (nums1[i - 1] == nums2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1] + 1;
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ans = Math.max(ans, dp[i][j]);
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int findLength(vector<int>& nums1, vector<int>& nums2) {
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int m = nums1.size(), n = nums2.size();
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vector<vector<int>> dp(m + 1, vector<int>(n + 1));
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int ans = 0;
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for (int i = 1; i <= m; ++i)
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{
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for (int j = 1; j <= n; ++j)
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{
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if (nums1[i - 1] == nums2[j - 1])
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{
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dp[i][j] = dp[i - 1][j - 1] + 1;
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ans = max(ans, dp[i][j]);
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func findLength(nums1 []int, nums2 []int) int {
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m, n := len(nums1), len(nums2)
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dp := make([][]int, m+1)
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for i := range dp {
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dp[i] = make([]int, n+1)
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}
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ans := 0
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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if nums1[i-1] == nums2[j-1] {
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dp[i][j] = dp[i-1][j-1] + 1
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if ans < dp[i][j] {
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ans = dp[i][j]
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}
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}
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}
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}
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return ans
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}
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```
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### **...**

‎solution/0700-0799/0718.Maximum Length of Repeated Subarray/README_EN.md

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@@ -37,13 +37,88 @@
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### **Python3**
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```python
40-
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class Solution:
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def findLength(self, nums1: List[int], nums2: List[int]) -> int:
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m, n = len(nums1), len(nums2)
43+
dp = [[0] * (n + 1) for _ in range(m + 1)]
44+
ans = 0
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for i in range(1, m + 1):
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for j in range(1, n + 1):
47+
if nums1[i - 1] == nums2[j - 1]:
48+
dp[i][j] = 1 + dp[i - 1][j - 1]
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ans = max(ans, dp[i][j])
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return ans
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```
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### **Java**
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```java
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class Solution {
57+
public int findLength(int[] nums1, int[] nums2) {
58+
int m = nums1.length;
59+
int n = nums2.length;
60+
int[][] dp = new int[m + 1][n + 1];
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int ans = 0;
62+
for (int i = 1; i <= m; ++i) {
63+
for (int j = 1; j <= n; ++j) {
64+
if (nums1[i - 1] == nums2[j - 1]) {
65+
dp[i][j] = dp[i - 1][j - 1] + 1;
66+
ans = Math.max(ans, dp[i][j]);
67+
}
68+
}
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}
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return ans;
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}
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}
73+
```
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### **C++**
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```cpp
78+
class Solution {
79+
public:
80+
int findLength(vector<int>& nums1, vector<int>& nums2) {
81+
int m = nums1.size(), n = nums2.size();
82+
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
83+
int ans = 0;
84+
for (int i = 1; i <= m; ++i)
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{
86+
for (int j = 1; j <= n; ++j)
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{
88+
if (nums1[i - 1] == nums2[j - 1])
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{
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dp[i][j] = dp[i - 1][j - 1] + 1;
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ans = max(ans, dp[i][j]);
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func findLength(nums1 []int, nums2 []int) int {
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m, n := len(nums1), len(nums2)
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dp := make([][]int, m+1)
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for i := range dp {
107+
dp[i] = make([]int, n+1)
108+
}
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ans := 0
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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if nums1[i-1] == nums2[j-1] {
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dp[i][j] = dp[i-1][j-1] + 1
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if ans < dp[i][j] {
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ans = dp[i][j]
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}
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}
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}
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}
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return ans
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}
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```
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49124
### **...**

‎solution/0700-0799/0718.Maximum Length of Repeated Subarray/Solution.cpp

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class Solution {
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public:
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int findLength(vector<int>& nums1, vector<int>& nums2) {
4+
int m = nums1.size(), n = nums2.size();
5+
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
6+
int ans = 0;
7+
for (int i = 1; i <= m; ++i)
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{
9+
for (int j = 1; j <= n; ++j)
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{
11+
if (nums1[i - 1] == nums2[j - 1])
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{
13+
dp[i][j] = dp[i - 1][j - 1] + 1;
14+
ans = max(ans, dp[i][j]);
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}
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}
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}
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return ans;
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}
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};

‎solution/0700-0799/0718.Maximum Length of Repeated Subarray/Solution.go

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1+
func findLength(nums1 []int, nums2 []int) int {
2+
m, n := len(nums1), len(nums2)
3+
dp := make([][]int, m+1)
4+
for i := range dp {
5+
dp[i] = make([]int, n+1)
6+
}
7+
ans := 0
8+
for i := 1; i <= m; i++ {
9+
for j := 1; j <= n; j++ {
10+
if nums1[i-1] == nums2[j-1] {
11+
dp[i][j] = dp[i-1][j-1] + 1
12+
if ans < dp[i][j] {
13+
ans = dp[i][j]
14+
}
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}
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}
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}
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return ans
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}

‎solution/0700-0799/0718.Maximum Length of Repeated Subarray/Solution.java

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@@ -1,14 +1,14 @@
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class Solution {
2-
public int findLength(int[] A, int[] B) {
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public int findLength(int[] nums1, int[] nums2) {
3+
int m = nums1.length;
4+
int n = nums2.length;
5+
int[][] dp = new int[m + 1][n + 1];
36
int ans = 0;
4-
int[][] dp = new int[A.length + 1][B.length + 1];
5-
for (int i = 1;i <= A.length;i++) {
6-
for (int j = 1;j <= B.length;j++) {
7-
if (A[i - 1] == B[j - 1]) {
7+
for (int i = 1; i <= m; ++i) {
8+
for (int j = 1; j <= n; ++j) {
9+
if (nums1[i - 1] == nums2[j - 1]) {
810
dp[i][j] = dp[i - 1][j - 1] + 1;
911
ans = Math.max(ans, dp[i][j]);
10-
} else {
11-
dp[i][j] = 0;
1212
}
1313
}
1414
}

‎solution/0700-0799/0718.Maximum Length of Repeated Subarray/Solution.py

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class Solution:
2+
def findLength(self, nums1: List[int], nums2: List[int]) -> int:
3+
m, n = len(nums1), len(nums2)
4+
dp = [[0] * (n + 1) for _ in range(m + 1)]
5+
ans = 0
6+
for i in range(1, m + 1):
7+
for j in range(1, n + 1):
8+
if nums1[i - 1] == nums2[j - 1]:
9+
dp[i][j] = 1 + dp[i - 1][j - 1]
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ans = max(ans, dp[i][j])
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return ans

‎solution/2000-2099/2070.Most Beautiful Item for Each Query/README.md

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@@ -72,7 +72,6 @@
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class Solution:
7373
def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:
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items.sort()
75-
n = len(items)
7675
prices = [p for p, _ in items]
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mx = [items[0][1]]
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for _, b in items[1:]:

‎solution/2000-2099/2070.Most Beautiful Item for Each Query/README_EN.md

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@@ -65,7 +65,6 @@ Hence, the answer to the query is 0.
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class Solution:
6666
def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:
6767
items.sort()
68-
n = len(items)
6968
prices = [p for p, _ in items]
7069
mx = [items[0][1]]
7170
for _, b in items[1:]:

‎solution/2000-2099/2070.Most Beautiful Item for Each Query/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -1,7 +1,6 @@
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class Solution:
22
def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:
33
items.sort()
4-
n = len(items)
54
prices = [p for p, _ in items]
65
mx = [items[0][1]]
76
for _, b in items[1:]:

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