[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: 2
Explanation: 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.

ぶんせき
2つの配列を与えて、同じ数をオンラインにして、交差できないことを要求して、私はあれこれ考えていなかったので、他の人のdpの考え方を参考にして、dp[i][j]は配列Aの前のiつの要素と配列Bの前のjつの要素の最も大連の線数を代表して、それからこの問題は最も長い公共のサブシーケンスを求める問題になりました.dp法は初期化と境界問題に注意すればよい.
コード#コード#

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