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Commit 6e67b4a

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feat: add solutions to lc problem: No.1458
No.1458.Max Dot Product of Two Subsequences
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‎solution/1400-1499/1458.Max Dot Product of Two Subsequences/README.md‎

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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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定义 $dp[i][j]$ 表示 $nums1$ 前 $i$ 个元素和 $nums2$ 前 $j$ 个元素得到的最大点积。
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那么有:
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$$
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dp[i][j]=max(dp[i-1][j], dp[i][j - 1], max(dp[i - 1][j - 1], 0) + nums1[i] \times nums2[j])
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$$
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答案为 $dp[m][n]$。
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时间复杂度 $O(m \times n),ドル空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是数组 $nums1$ 和 $nums2$ 的长度。
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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 maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
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m, n = len(nums1), len(nums2)
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dp = [[-inf] * (n + 1) for _ 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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v = nums1[i - 1] * nums2[j - 1]
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1],
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max(dp[i - 1][j - 1], 0) + v)
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return dp[-1][-1]
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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 maxDotProduct(int[] nums1, int[] nums2) {
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int m = nums1.length, n = nums2.length;
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int[][] dp = new int[m + 1][n + 1];
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for (int[] e : dp) {
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Arrays.fill(e, Integer.MIN_VALUE);
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}
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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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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = Math.max(dp[i][j], Math.max(0, dp[i - 1][j - 1]) + nums1[i - 1] * nums2[j - 1]);
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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 maxDotProduct(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, INT_MIN));
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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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int v = nums1[i - 1] * nums2[j - 1];
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = max(dp[i][j], max(0, dp[i - 1][j - 1]) + v);
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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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### **Go**
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```go
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func maxDotProduct(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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for j := range dp[i] {
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dp[i][j] = math.MinInt32
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}
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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v := nums1[i-1] * nums2[j-1]
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dp[i][j] = max(dp[i-1][j], dp[i][j-1])
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dp[i][j] = max(dp[i][j], max(0, dp[i-1][j-1])+v)
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}
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}
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return dp[m][n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1400-1499/1458.Max Dot Product of Two Subsequences/README_EN.md‎

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## Solutions
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Dynamic Programming.
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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 maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
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m, n = len(nums1), len(nums2)
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dp = [[-inf] * (n + 1) for _ 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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v = nums1[i - 1] * nums2[j - 1]
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1],
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max(dp[i - 1][j - 1], 0) + v)
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return dp[-1][-1]
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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 maxDotProduct(int[] nums1, int[] nums2) {
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int m = nums1.length, n = nums2.length;
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int[][] dp = new int[m + 1][n + 1];
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for (int[] e : dp) {
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Arrays.fill(e, Integer.MIN_VALUE);
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}
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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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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = Math.max(dp[i][j], Math.max(0, dp[i - 1][j - 1]) + nums1[i - 1] * nums2[j - 1]);
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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 maxDotProduct(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, INT_MIN));
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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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int v = nums1[i - 1] * nums2[j - 1];
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = max(dp[i][j], max(0, dp[i - 1][j - 1]) + v);
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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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### **Go**
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```go
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func maxDotProduct(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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for j := range dp[i] {
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dp[i][j] = math.MinInt32
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}
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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v := nums1[i-1] * nums2[j-1]
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dp[i][j] = max(dp[i-1][j], dp[i][j-1])
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dp[i][j] = max(dp[i][j], max(0, dp[i-1][j-1])+v)
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}
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}
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return dp[m][n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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class Solution {
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public:
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int maxDotProduct(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, INT_MIN));
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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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int v = nums1[i - 1] * nums2[j - 1];
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = max(dp[i][j], max(0, dp[i - 1][j - 1]) + v);
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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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func maxDotProduct(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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for j := range dp[i] {
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dp[i][j] = math.MinInt32
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}
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}
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for i := 1; i <= m; i++ {
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for j := 1; j <= n; j++ {
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v := nums1[i-1] * nums2[j-1]
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dp[i][j] = max(dp[i-1][j], dp[i][j-1])
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dp[i][j] = max(dp[i][j], max(0, dp[i-1][j-1])+v)
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}
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}
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return dp[m][n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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class Solution {
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public int maxDotProduct(int[] nums1, int[] nums2) {
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int m = nums1.length, n = nums2.length;
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int[][] dp = new int[m + 1][n + 1];
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for (int[] e : dp) {
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Arrays.fill(e, Integer.MIN_VALUE);
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}
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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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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = Math.max(dp[i][j], Math.max(0, dp[i - 1][j - 1]) + nums1[i - 1] * nums2[j - 1]);
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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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class Solution:
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def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
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m, n = len(nums1), len(nums2)
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dp = [[-inf] * (n + 1) for _ 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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v = nums1[i - 1] * nums2[j - 1]
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dp[i][j] = max(dp[i - 1][j], dp[i][j - 1],
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max(dp[i - 1][j - 1], 0) + v)
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return dp[-1][-1]

‎solution/1400-1499/1461.Check If a String Contains All Binary Codes of Size K/README.md‎

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遍历字符串 $s,ドル用一个哈希表存储所有长度为 $k$ 的不同子串。只需要判断子串数能否达到 2ドル^k$ 即可。
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时间复杂度 $O(n \times k),ドル其中 $n$ 是字符串 $s$ 的长度,$k$ 是子串长度。
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时间复杂度 $O(n \times k),ドル其中 $n$ 是字符串 $s$ 的长度,$k$ 是子串长度。
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**方法二:滑动窗口**
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