动态规划May-25th “Uncrossed Lines (Python3)”

Posted 2021-04-06 迪乐阅读

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“ 五月，总结模板的日子.”

Day 25  ——  Uncrossed Lines

不交叉的线

问题描述

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. 1 <= A.length <= 500

2. 1 <= B.length <= 500

3. 1 <= A[i], B[i] <= 2000

解题思路  

【动态规划 】

递推公式：

       设 A[0] ~ A[x] 与 B[0] ~ B[y] 的最大连线数为 f(x, y)，那么对于任意位置的 f(i, j) 而言：

        从后往前遍历，可以得到 递推公式：

• 如果A[i] == B[j] -> dp_table[i][j] = dp_table[i+1][j+1] + 1
• 否则 -> dp_table[i][j] = max(dp_table[i+1][j], dp_table[i][j+1])

最后dp[0][0]就是最大的划线数量

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