代码随想录算法训练营Day53 | 动态规划(14/17) LeetCode 1143.最长公共子序列 1035.不相交的线

继续练习子序列的问题

第一题

1143. Longest Common Subsequence

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

• For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

本题和动态规划：LC 718的区别在于这里不要求是连续的，其他思路都一样的

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        dp = [[0] * (len(text2) + 1) for _ in range(len(text1) + 1)]
        for i in range(1, len(text1) + 1):
            for j in range(1, len(text2) + 1):
                if text1[i - 1] == text2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

        return dp[len(text1)][len(text2)]

第二题

1035. Uncrossed Lines

You are given two integer arrays nums1 and nums2. We write the integers of nums1 and nums2 (in the order they are given) on two separate horizontal lines.

We may draw connecting lines: a straight line connecting two numbers nums1[i] and nums2[j] such that:

• nums1[i] == nums2[j], and
• the line we draw does not intersect any other connecting (non-horizontal) line.

Note that a connecting line cannot intersect even at the endpoints (i.e., each number can only belong to one connecting line).

Return the maximum number of connecting lines we can draw in this way.

直线不能相交，这就是说明在字符串A中 找到一个与字符串B相同的子序列，且这个子序列不能改变相对顺序，只要相对顺序不改变，链接相同数字的直线就不会相交。

本题说是求绘制的最大连线数，其实就是求两个字符串的最长公共子序列的长度！

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