1035. Uncrossed Lines

最新推荐文章于 2025-02-16 22:09:54 发布

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最新推荐文章于 2025-02-16 22:09:54 发布

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分类专栏： leetcode

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本文探讨了如何在两个整数列表A和B中找到最大数量的不相交连线，这些连线连接相同数值的元素，且不与其他连线交叉。通过使用动态规划方法，我们能够有效地解决这个问题，并给出了详细的算法实现。

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

Now, we may draw a straight line connecting two numbers A[i] and B[j] as long as A[i] == B[j], and the line we draw does not intersect any other connecting (non-horizontal) 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

思路：setting和最长公共子串一样，不能走回头路

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