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maolonglongmaolonglong
committedMay 21, 2021
feat: add solutions to lc problem: No.1035. Uncrossed Lines
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‎solution/1000-1099/1035.Uncrossed Lines/README.md

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<!-- 这里可写通用的实现逻辑 -->
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最长公共子序列问题
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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 maxUncrossedLines(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 i in range(m + 1)]
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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] = dp[i - 1][j - 1] + 1
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else:
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
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return dp[m][n]
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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 maxUncrossedLines(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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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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} else {
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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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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 maxUncrossedLines(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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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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} else {
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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}
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};
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```
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### **...**

‎solution/1000-1099/1035.Uncrossed Lines/README_EN.md

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@@ -101,18 +101,68 @@ We cannot draw 3 uncrossed lines, because the line from A[1]=4 to B[2]=4 will in
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## Solutions
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Longest common sub-sequences
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def maxUncrossedLines(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 i in range(m + 1)]
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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] = dp[i - 1][j - 1] + 1
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else:
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
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return dp[m][n]
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```
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### **Java**
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```java
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class Solution {
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public int maxUncrossedLines(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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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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} else {
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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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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 maxUncrossedLines(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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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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} else {
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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}
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};
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```
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### **...**

‎solution/1000-1099/1035.Uncrossed Lines/Solution.cpp

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class Solution {
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public:
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int maxUncrossedLines(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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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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} else {
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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}
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};

‎solution/1000-1099/1035.Uncrossed Lines/Solution.java

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class Solution {
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public int maxUncrossedLines(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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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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} else {
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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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return dp[m][n];
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}
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}

‎solution/1000-1099/1035.Uncrossed Lines/Solution.py

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class Solution:
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def maxUncrossedLines(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 i in range(m + 1)]
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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] = dp[i - 1][j - 1] + 1
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else:
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
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return dp[m][n]

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