LeetCode 1035. Uncrossed Lines

Posted on | In |
Words count in article: 339 | Reading time ≈ 2

题目

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:

1
2
3
4
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:

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

Example 3:

1
2
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

思路

DP.

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution(object):
    def maxUncrossedLines(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: int
        """
        dp = [[0 for _ in range(len(B))] for _ in range(len(A))]
        for i in range(len(A)):
            for j in range(len(B)):
                if A[i] == B[j]:
                    dp[i][j] = max(dp[i][j], 1)
                    if i-1 >= 0 and j-1 >= 0:
                        dp[i][j] = dp[i-1][j-1]+1
                else:
                    if i-1 >= 0 and j-1 >= 0:
                        dp[i][j] = max(dp[i][j], dp[i-1][j-1])
                    if i-1 >= 0:
                        dp[i][j] = max(dp[i][j], dp[i-1][j])
                    if j-1 >= 0:
                        dp[i][j] = max(dp[i][j], dp[i][j-1])
        return dp[-1][-1]