# 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

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


dp[0][0] = 0
dp[i][j]就是最后结果



### 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];
}
}