【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.

我們在兩條不同的平行線上寫下分別(按照給定的順序)寫下數列A和B。

現在我們再加上連接線：一條直線在A[i]和B[j]上，當：

• A[i] == B[j];
• 這條線不會和其他的連接線相交。

注意，即使只有在頂點連接也算是相交，也就是說，每個數字只能被一條連接線給使用。

回傳在以上條件，我們能夠畫出的連接線可能的最大數值。

Example:

筆者：WTF 這圖也太大…

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

Solution

• 仔細想想之後，便可以發現這題就是LCS(longest common subsequence)，因此可以用DP解。
  • 同樣都可以想成，要刪掉某些數字/字母之後，讓兩個輸入變成一樣的，然後求長度。
• 將問題切割成子問題，也就是將連線的部分和剩餘的部分分開。
• 其遞迴式為：
  
  $$\begin{gathered} f[1][1]=same(1,1) \\ f[i][j]=max\{f[i-1][j-1]+same(i,j),f[i-1][j],f[i][j-1]\} \end{gathered}$$

Code

class Solution {
public:
    int maxUncrossedLines(vector<int>& A, vector<int>& B) {
        int sA = A.size(), sB = B.size();
        vector<vector<int>> dp(sA + 1, vector<int>(sB + 1, 0));
        
        for(int i = 0; i < sA; i++)
            for(int j = 0; j < sB; j++)
                if(A[i] == B[j])
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                else
                    dp[i + 1][j + 1] = max(dp[i + 1][j], dp[i][j + 1]);
        
        return dp[sA][sB];
    }
};

tags: LeetCode C++