update: 添加问题“1458.两个子序列的最大点积”的代码和题解

.commitmsg

@@ -0,0 +1,7 @@
+1458: AC.cpp (#1310) + word(en) + fix(pic)
+
+cpp - AC,84.23%,29.53%
+py - AC,97.99%,39.85%
+go - AC,76.92%,38.46%
+java - AC,65.24%,32.38%
+rust - AC,100.00%,60.00%

Codes/1458-max-dot-product-of-two-subsequences.cpp

@@ -0,0 +1,32 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-01-08 09:21:01
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-01-08 09:23:30
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
+        int n = nums1.size(), m = nums2.size();
+        vector<vector<int>> dp(n, vector<int>(m));
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                dp[i][j] = nums1[i] * nums2[j];
+                if (i) {
+                    dp[i][j] = max(dp[i][j], dp[i - 1][j]);
+                }
+                if (j) {
+                    dp[i][j] = max(dp[i][j], dp[i][j - 1]);
+                }
+                if (i && j) {
+                    dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + nums1[i] * nums2[j]);
+                }
+            }
+        }
+        return dp[n - 1][m - 1];
+    }
+};

Codes/1458-max-dot-product-of-two-subsequences.go

@@ -0,0 +1,36 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-01-08 09:21:01
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-01-08 09:36:59
+ */
+package main
+
+func max1458_2(a, b int) int {
+    if a > b {
+        return a
+    }
+    return b
+}
+
+func max1458(a, b, c, d int) int {
+    return max1458_2(a, max1458_2(b, max1458_2(c, d)))
+}
+
+func maxDotProduct(nums1 []int, nums2 []int) int {
+    n, m := len(nums1), len(nums2)
+    dp := make([][]int, n + 1)
+    for i := range dp {
+        dp[i] = make([]int, m + 1)
+        for j := range dp[i] {
+            dp[i][j] = -1000000;
+        }
+    }
+
+    for i, x := range nums1 {
+        for j, y := range nums2 {
+            dp[i+1][j+1] = max1458(x*y, dp[i][j+1], dp[i+1][j], dp[i][j] + x*y)
+        }
+    }
+    return dp[n][m]  // 不是dp[n-1][m-1]
+}

Codes/1458-max-dot-product-of-two-subsequences.java

@@ -0,0 +1,23 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-01-08 09:21:01
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-01-08 09:45:31
+ */
+import java.util.Arrays;
+
+class Solution {
+    public int maxDotProduct(int[] nums1, int[] nums2) {
+        int n = nums1.length, m = nums2.length;
+        int[][] dp = new int[n+1][m+1];
+        for (int i = 0; i <= n; i++) {
+            Arrays.fill(dp[i], -1000000);
+        }
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= m; j++) {
+                dp[i][j] = Math.max(nums1[i-1] * nums2[j-1], Math.max(dp[i-1][j], Math.max(dp[i][j - 1], dp[i-1][j-1] + nums1[i-1] * nums2[j-1])));
+            }
+        }
+        return dp[n][m];
+    }
+}

Codes/1458-max-dot-product-of-two-subsequences.py

@@ -0,0 +1,15 @@
+'''
+Author: LetMeFly
+Date: 2026-01-08 09:21:01
+LastEditors: LetMeFly.xyz
+LastEditTime: 2026-01-08 09:29:41
+'''
+from typing import List
+
+class Solution:
+    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
+        dp = [[-1000000000] * (len(nums2) + 1) for _ in range(len(nums1) + 1)]
+        for i, x in enumerate(nums1, 1):
+            for j, y in enumerate(nums2, 1):
+                dp[i][j] = max(x * y, dp[i-1][j], dp[i][j-1], dp[i-1][j-1] + x * y)
+        return dp[-1][-1]

Codes/1458-max-dot-product-of-two-subsequences.rs

@@ -0,0 +1,19 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-01-08 09:21:01
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-01-08 09:51:58
+ */
+impl Solution {
+    pub fn max_dot_product(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
+        let n: usize = nums1.len();
+        let m: usize = nums2.len();
+        let mut dp: Vec<Vec<i32>> = vec![vec![-1000000; m+1]; n+1];
+        for i in 1..=n {
+            for j in 1..=m {
+                dp[i][j] = dp[i-1][j].max(dp[i][j-1].max((nums1[i-1] * nums2[j-1]).max(dp[i-1][j-1] + nums1[i-1] * nums2[j-1])));
+            }
+        }
+        dp[n][m]
+    }
+}

Codes/lib.rs

@@ -5,7 +5,7 @@
  * @LastEditTime: 2026-01-06 13:17:16
  */
 pub struct Solution;
-include!("1161-maximum-level-sum-of-a-binary-tree_20260106.rs");  // 这个fileName是会被脚本替换掉的
+include!("1458-max-dot-product-of-two-subsequences.rs");  // 这个fileName是会被脚本替换掉的
 
 #[derive(Debug, PartialEq, Eq)]
 pub struct TreeNode {

README.md

@@ -576,6 +576,7 @@
 |1450.在既定时间做作业的学生人数|简单|<a href="https://leetcode.cn/problems/number-of-students-doing-homework-at-a-given-time/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/08/19/LeetCode%201450.%E5%9C%A8%E6%97%A2%E5%AE%9A%E6%97%B6%E9%97%B4%E5%81%9A%E4%BD%9C%E4%B8%9A%E7%9A%84%E5%AD%A6%E7%94%9F%E4%BA%BA%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/126417256" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-students-doing-homework-at-a-given-time/solution/letmefly-1450zai-ji-ding-shi-jian-zuo-zu-0sd7/" target="_blank">LeetCode题解</a>|
 |1455.检查单词是否为句中其他单词的前缀|简单|<a href="https://leetcode.cn/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/08/21/LeetCode%201455.%E6%A3%80%E6%9F%A5%E5%8D%95%E8%AF%8D%E6%98%AF%E5%90%A6%E4%B8%BA%E5%8F%A5%E4%B8%AD%E5%85%B6%E4%BB%96%E5%8D%95%E8%AF%8D%E7%9A%84%E5%89%8D%E7%BC%80/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/126448124" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/solution/letmefly-1455jian-cha-dan-ci-shi-fou-wei-m3rs/" target="_blank">LeetCode题解</a>|
 |1457.二叉树中的伪回文路径|中等|<a href="https://leetcode.cn/problems/pseudo-palindromic-paths-in-a-binary-tree/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/11/25/LeetCode%201457.%E4%BA%8C%E5%8F%89%E6%A0%91%E4%B8%AD%E7%9A%84%E4%BC%AA%E5%9B%9E%E6%96%87%E8%B7%AF%E5%BE%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134617854" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/pseudo-palindromic-paths-in-a-binary-tree/solutions/2541535/letmefly-1457er-cha-shu-zhong-de-wei-hui-91vp/" target="_blank">LeetCode题解</a>|
+|1458.两个子序列的最大点积|困难|<a href="https://leetcode.cn/problems/max-dot-product-of-two-subsequences/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/01/08/LeetCode%201458.%E4%B8%A4%E4%B8%AA%E5%AD%90%E5%BA%8F%E5%88%97%E7%9A%84%E6%9C%80%E5%A4%A7%E7%82%B9%E7%A7%AF/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/156713484" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/max-dot-product-of-two-subsequences/solutions/3875454/letmefly-1458liang-ge-zi-xu-lie-de-zui-d-3f54/" target="_blank">LeetCode题解</a>|
 |1460.通过翻转子数组使两个数组相等|简单|<a href="https://leetcode.cn/problems/make-two-arrays-equal-by-reversing-sub-arrays/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/08/24/LeetCode%201460.%E9%80%9A%E8%BF%87%E7%BF%BB%E8%BD%AC%E5%AD%90%E6%95%B0%E7%BB%84%E4%BD%BF%E4%B8%A4%E4%B8%AA%E6%95%B0%E7%BB%84%E7%9B%B8%E7%AD%89/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/126499790" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/make-two-arrays-equal-by-reversing-sub-arrays/solution/letmefly-1460tong-guo-fan-zhuan-zi-shu-z-q812/" target="_blank">LeetCode题解</a>|
 |1462.课程表IV|中等|<a href="https://leetcode.cn/problems/course-schedule-iv/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/09/12/LeetCode%201462.%E8%AF%BE%E7%A8%8B%E8%A1%A8IV/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132825649" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/course-schedule-iv/solutions/2438404/letmefly-1462ke-cheng-biao-ivtuo-bu-pai-gufke/" target="_blank">LeetCode题解</a>|
 |1464.数组中两元素的最大乘积|简单|<a href="https://leetcode.cn/problems/maximum-product-of-two-elements-in-an-array/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/08/26/LeetCode%201464.%E6%95%B0%E7%BB%84%E4%B8%AD%E4%B8%A4%E5%85%83%E7%B4%A0%E7%9A%84%E6%9C%80%E5%A4%A7%E4%B9%98%E7%A7%AF/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/126536351" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-product-of-two-elements-in-an-array/solution/letmefly-1464shu-zu-zhong-liang-yuan-su-txf8l/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 1411.给Nx3网格图涂色的方案数.md

@@ -3,7 +3,7 @@ title: 1411.给 N x 3 网格图涂色的方案数：递推+原地滚动(动态
 date: 2026-01-03 21:46:12
 tags: [题解, LeetCode, 困难, 动态规划, DP]
 categories: [题解, LeetCode]
-index_img: https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/04/12/e1.png
+index_img: https://assets.leetcode.cn/aliyun-lc-upload/uploads/2020/04/12/e1.png
 ---
 
 # 【LetMeFly】1411.给 N x 3 网格图涂色的方案数：递推+原地滚动(动态规划)
@@ -23,7 +23,7 @@ index_img: https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/04/12/e1
 <pre><strong>输入：</strong>n = 1
 <strong>输出：</strong>12
 <strong>解释：</strong>总共有 12 种可行的方法：
-<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/04/12/e1.png" style="height: 289px; width: 450px;">
+<img alt="" src="https://assets.leetcode.cn/aliyun-lc-upload/uploads/2020/04/12/e1.png" style="height: 289px; width: 450px;">
 </pre>
 
 <p><strong>示例 2：</strong></p>