LeetCode解题分享：1143. Longest Common Subsequence

Problem

Given two strings text1 and text2, return the length of their longest common subsequence.

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. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

Example 1:

Input: text1 = “abcde”, text2 = “ace”
Output: 3
Explanation: The longest common subsequence is “ace” and its length is 3.

Example 2:

Input: text1 = “abc”, text2 = “abc”
Output: 3
Explanation: The longest common subsequence is “abc” and its length is 3.

Example 3:

Input: text1 = “abc”, text2 = “def”
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Constraints:

1. 1 <= text1.length <= 1000
2. 1 <= text2.length <= 1000
3. The input strings consist of lowercase English characters only.

解题思路

   这是一道很典型的动态规划题目。对于这种关于计算两个字符串之间的关系的题，大多都是定义一个二维数组， 数组中的每个元素都表示一个状态。对于这一题，我们可以定义一个二维数组$dp[i][j]$，数组中的每一个元素表示当前状态（即text1的前$i$个字符和text2的前$j$个字符可以计算得出的最长子字符串，需要注意的是这里的$i$和$j$并不包括在内。）下最长相同子字符串的长度， 那么下一步就应该分析状态转移方程。

   很明显，当我们扫描到text1的第$i$个字符，text2的第$j$个字符时，需要考虑这两个字符是否相等，如果这两个字符相等，很明显，计算出的结果应该是text1的前$i-1$和text2的前$j-1$个计算出的最长子字符串的长度加1，即：
$$dp[i][j]=dp[i-1][j-1]+1$$

   当两个字符不相等时，我们就需要考虑$dp[i-1][j]$以及$dp[i][j-1]$，因为当前的两个字符已经不相等了，所以我们没有办法在原有的基础上更进一步，只能在$dp[i-1][j]$以及$dp[i][j-1]$中选择一个最大值，即：
$$dp[i][j]=max(dp[i-1][j],dp[i][j-1])$$

   代码如下：

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

        for i in range(1, rows):
            for j in range(1, cols):
                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[-1][-1]