LeetCode-1035. Uncrossed Lines

LeetCode-1035. Uncrossed Lines

Posted by Jae on October 29, 2019

1、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:

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

2、思路

该题意为将上下两行数组中相等元素使用不交叉的直线相连,问最多有多少条这样的直线?这道题我们可使用动态规划的思想求解:

确定状态:

题目要求数组A[a1,a2…ai]和数组B[b1,b2..bj]最多有多少条不交叉的连线,我们假设最优策略中ai==bj,也就是ai和bj相连,此时转化成子问题: 求数组A[a1,a2…ai-1]和数组B[b1,b2…bj-1]最多有多少条不交叉的连线,所以我们定义dp[i][j]表示数组A[a1,a2…ai]和数组B[b1,b2…bj]最多有多少条不交叉的连线。

转移方程:

设状态dp[i][j]=数组A[a1..ai]和数组B[b1…bj]最多有多少条不交叉的连线,对于任意的i,j都有:

dp[i][j] = dp[i-1][j-1]+1 当A[i]=B[j]时
dp[i][j] = max{dp[i-1][j], dp[i][j-1]} 当A[i] != B[j]时

初始条件和边界情况

dp[0][0] = 0
dp[i][j]就是最后结果
算法时间复杂度为O(mn)m n分别为数组A和B的长度
空间复杂度为O(mn)

3 实现

public class Solution {
    public int maxUncrossedLines(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        int[][] dp = new int[m+1][n+1];

        // 初始化
        dp[0][0] = 0;
        for (int i = 1; i <= m; i++){
            for (int j = 1; j <= n; j++){
                if (A[i-1] == B[j-1]){
                    dp[i][j] = dp[i-1][j-1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        return dp[m][n];
    }
}