Leetcode 1035. Uncrossed Lines

Problem

You are given two integer arrays nums1 and nums2. We write the integers of nums1 and nums2 (in the order they are given) on two separate horizontal lines.

We may draw connecting lines: a straight line connecting two numbers nums1i and nums2j such that:

  • nums1i == nums2j, and
  • the line we draw does not intersect any other connecting (non-horizontal) line.

Note that a connecting line cannot intersect even at the endpoints (i.e., each number can only belong to one connecting line).

Return the maximum number of connecting lines we can draw in this way.

Algorithm

Dynamic Programming (DP): same as Longest Common Subsequence (LCS).

  • If s1[i] != s2[j]:
    F ( i , j ) = max ⁡ ( F ( i − 1 , j ) , F ( i , j − 1 ) ) F(i, j) = \max\left( F(i-1, j), F(i, j-1) \right) F(i,j)=max(F(i−1,j),F(i,j−1))

  • If s1[i] == s2[j]:
    F ( i , j ) = F ( i − 1 , j − 1 ) + 1 F(i, j) = F(i-1, j-1) + 1 F(i,j)=F(i−1,j−1)+1

Code

python3 复制代码
class Solution:
    def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:
        l1, l2 = len(nums1) + 1, len(nums2) + 1
        dp = [[0] * l2 for _ in range(l1)] 

        for i in range(1, l1):
            for j in range(1, l2):
                if nums1[i-1] == nums2[j-1]:
                    dp[i][j] = dp[i-1][j-1] + 1
                else:
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1])
        
        return dp[l1-1][l2-1]
相关推荐
布莱克6057 小时前
C++ 中 map 的 insert 与 emplace:key 已存在时会覆盖吗?
开发语言·数据结构·c++·算法·map
D e re k ZH10 小时前
C++ 系统学习手记:从语法细节到工程实践
开发语言·数据结构·c++·算法
石小石Orz10 小时前
AI 正在蚕食我们的编码能力
前端·后端·算法
民乐团扒谱机12 小时前
【微实验】NSGA-II(非支配排序遗传算法)——解决多目标优化问题的“自然选择模拟器”
linux·人工智能·算法
青少儿编程课堂12 小时前
蓝桥杯·全国青少年信息素养大赛·电子学会Python/C++考级:中缀表达式求值解析
c++·python·算法·bfs·信息学竞赛
文丑颜不良啊13 小时前
LeetCode2261~2280题目解析
java·leetcode
武子康14 小时前
Qwen-Drive 到底是什么:4B 级 VLM 如何成为驾驶策略底座
人工智能·算法·机器学习
zh_xuan16 小时前
LeeCode 496. 下一个更大元素I
数据结构·算法·栈·单调栈
机 _ 长16 小时前
05 智算运维|8 张卡怎么才算「捆成一张」:NCCL 实测与拓扑
运维·人工智能·算法·智算运维