[leetcode] 1035. Uncrossed Lines

[leetcode] 1035. Uncrossed Lines

Description

We write the integers of A and B (in the order they are given) on two separate horizontal lines.

Now, we may draw connecting lines: a straight line connecting two numbers A[i] and B[j] such that:

A[i] == B[j];The line we draw does not intersect any other connecting (non-horizontal) line.

Note that a connecting lines cannot intersect even at the endpoints: each number can only belong to one connecting line.

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

Example 1:

Input: A = [1,4,2], B = [1,2,4]Output: 2Explanation: We can draw 2 uncrossed lines as in the diagram.We cannot draw 3 uncrossed lines, because the line from A[1]=4 to B[2]=4 will intersect the line from A[2]=2 to B[1]=2.

Example 2:

Input: A = [2,5,1,2,5], B = [10,5,2,1,5,2]Output: 3

Example 3:

Input: A = [1,3,7,1,7,5], B = [1,9,2,5,1]Output: 2

Note:

1 <= A.length <= 500.1 <= B.length <= 500.1 <= A[i], B[i] <= 2000.

分析

题目的意思是：给定两个数组，给相同的数连上线，要求不能够交叉，我左思右想没想出来，所以借鉴了一下别人的dp的思路，dp[i][j]代表数组A前i个元素和数组B前j个元素的最大连线数，然后这个问题就变成了类似求最长公共子序列的问题了，dp方法要注意初始化和边界问题就行了。

代码

class Solution: def maxUncrossedLines(self, A: List[int], B: List[int]) -> int: m=len(A) n=len(B) dp=[[0 for _ in range(n+1)] for _ in range(m+1)] for i in range(1,m+1): for j in range(1,n+1): if(A[i-1]==B[j-1]): dp[i][j]=dp[i-1][j-1]+1 else: dp[i][j]=max(dp[i][j-1],dp[i-1][j]) return dp[m][n]

参考文献

​​[LeetCode] Python by O( m*n ) DPw/ Graph​​

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